MAIN
JEEJEEMAIN
CO
CM
OP
MLPELTEET E
MATHEMATICS
MATHEMATICS
MAIN
JEEJEEMAIN
CO
CM
OP
MLPELTEET E
MATHEMATICS
MATHEMATICS
Ravi Prakash
Retired Associate Professor, Rajdhani College
University of Delhi, Delhi
Ajay Kumar
Professor of Mathematics
University of Delhi, Delhi
Usha Gupta
Retired Associate Professor, Rajdhani College
University of Delhi, Delhi
McGraw Hill Education (India) Private Limited
CHENNAI
McGraw Hill Education Offices
Chennai New York St Louis San Francisco Auckland Bogotá Caracas
Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal
San Juan Santiago Singapore Sydney Tokyo Toronto
Published by McGraw Hill Education (India) Private Limited,
444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai-600116
Complete Mathematics—JEE Main
Copyright © 2018, McGraw Hill Education (India) Private Limited.
No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program
listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication.
This edition can be exported from India only by the publishers,
McGraw Hill Education (India) Private Limited.
Price: `895/1 2 3 4 5 6 7 8 9 7085462 22 21 20 19 18
Printed and bound in India
ISBN (13): 978-93-87572-56-0
ISBN (10): 93-87572-56-0
Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable.
However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information
published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or
damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education
(India) and its authors are supplying information but are not attempting to render engineering or other professional services. If
such services are required, the assistance of an appropriate professional should be sought.
Typeset at Sri Krishna Graphics, Delhi and printed at
Cover Designer: Neeraj Dayal
visit us at: www.mheducation.co.in
To Our Readers…
How to Crack the JEE
This new edition of our book, Complete Mathematics JEE Main, is primarily meant for the candidates due to appear
in CBSE’s JEE (Main). It thoroughly covers all the topics prescribed in the Mathematics syllabus of this examination.
The book is divided into 28 chapters. Each chapter is organised as follows:
1. It begins with a review of important definitions, formulae, theorems, tips, tricks and techniques to solve
problems. Difficult concepts have been explained by giving appropriate illustrations in sufficient numbers.
2. This theory is followed by examples. These examples (solved in detail) are divided into three categories:
(a) Concept Based examples to further strengthen and enrich your conceptual understanding
(b) Level 1 examples carefully selected to groom you for the JEE Main examination. This section contains both
Straight Objective Type Questions and Assertion-Reason Type Questions.
(c) Level 2 examples added to further boost your confidence by initiating you into tougher questions
3. The Solved Examples are followed by Exercises. These have been divided into five parts.
(i) Concept Based Questions
(ii) Level 1 Questions
(iii) Level 2 Questions
(iv) Previous Years’ AIEEE/JEE (Main) Questions
(v) Previous Years’ B. Architecture Entrance Examination Questions.
The philosophy behind (i), (ii) and (iii) is the same as that in Solved Examples. We have added (iv) and (v) to
give you glimpse of the questions that have been asked in the previous years (till 2016). Questions from online
JEE (Main) papers from 2013 and 2016 have also been added. As the Mathematics syllabus for B. Architecture
Entrance Examination Questions is the same as that of JEE (Main) examination, questions from these papers have
also been added for further practice.
4. Each exercise has been provided with answers.
5. Hints and Solutions to all the questions in (i), (ii), (iii), (iv) and (v) have been given at the end of the chapters.
How to Use this Book
Step 1
Read the theory given in the chapter carefully. Take meaningful notes in a notebook.
Step 2
Go to solved examples. First attempt the questions given in this section (Concept-based). Try to solve each example
yourself. If you fail to do it within a reasonable time limit, go through the solution carefully.
The key to improvement is NOT merely understanding the explanation given in the solution, but to see exactly what
clues in the question lead us to the correct answer.
Step 3
Repeat Step 2 with Concept-based questions in the exercise.
Step 4
Repeat Step 2 and Step 3 with Solved Examples and questions in Exercises at Level-1 and Level-2.
Important: Leave Previous Years’ Questions for the future.
vi To Our Readers…
Facing the Examination
The analysis of previous fourteen years’ papers shows that it is not possible to leave any topic. You get questions from
each and every topic of mathematics. Thus, it is essential that you study each and every topic thoroughly.
Beginning the Preparation
The best time to begin preparation for JEE (Main) is when you have studied almost fifty percent of your Class XI course,
that is, almost one and a half year before the day you wish to appear in the JEE (Main).
Effective Studying
You need to set priorities so that you can get the best return for time invested. Write down the time you wish to devote
to Physics, Chemistry and Mathematics. As you study, you will develop a list of your problem areas, which will direct
you further. This may require you to rearrange your priorities. Do not overestimate the time you have, extra preparation
always helps. The key is to set up a workable plan and stick to it.
Some Tips for Preparation
∑ Identify your strengths and weaknesses.
It is important to identify your problem areas early on. Your study plan would need to evolve as you identify these
areas.
Equally important is to know your strength areas. You should ensure that these areas are further strengthened
and that you are not a victim of overconfidence.
Your study plan should strike a balance between the time kept for strength areas and for problem areas. Though
you need to concentrate more on your problem areas, mastering your strength areas in between would ensure that
you do not get dejected.
∑ Make notes and continually review each section.
Take meaningful notes in a notebook which will be of a great help at a later date. Good notes are short and crisp,
but cover the key concepts. Very long notes are redundant as you never have time to go through them and too
short notes are useless as they do not trigger any recall.
Rapidly review the section you have just studied. Recall all the major ideas of this portion. If you have any
difficulties, check your notes.
∑ Study for short durations.
Your brain needs time to assimilate your thoughts and develop a deep understanding of each concept that you are
learning. Try to study in short durations with rest breaks in between. During these rests, first blank out your mind
so as to relax and then go over the key concepts that you have learnt in the just completed leg of your study.
∑ Do each exercise keeping time in mind and look for clues.
This is very important. Before proceeding to another section, rework all the questions about which you were not
sure or which you could not solve correctly. And make sure that you complete them in the time allocated for each
section.
∑ Always have a positive attitude.
Be motivated. Don’t be discouraged if you are unable to solve a problem. When you cannot understand a problem
you tend to think in different directions. This helps you to develop your mental faculties.
∑ Always keep your goal in mind.
∑ Get your body clock in rhythm with the examination timings a few days before the examination.
Some students are in the habit of studying at night and sleeping during the day time. This habit should not be
continued till the day of the exam. You should develop a habit of working during the same clock hours as the
actual exam at least 15-20 days before the exam. This will help you to work effectively during the exam.
To Our Readers…
vii
Test Taking Strategies
∑ Read the instructions in the question paper carefully and scrupulously follow these instructions.
∑ Read the entire question before attempting to answer it, and recognise the key concepts in the same. Each question typically contains certain key concepts; certain pieces of information that, if you recognise accurately, your
ability to come to the right answer increases significantly. The correct answer will typically incorporate all the
key concepts contained in the question.
∑ Read each question carefully. Answer the questions asked, not the one you may have expected.
∑ Budget your time. Work swiftly. Don’t devote too much time to any difficult problem. There may be some simple
ones waiting for you. Concentrate on those questions which you are absolutely or reasonably certain you know.
Leave those questions for later where you have some doubts. Go through the exam answering questions that you
can answer and skip questions you can’t answer. Then go through the paper again, deliberating and answering
the questions you had left unattended earlier.
Best wishes for your endeavours!
Ravi Prakash
Ajay Kumar
Usha Gupta
Please log on to www.mhhe.com/jeemaincompletemathematics for Seven Sectional Test
Papers, Five Practice Test Papers and Miscellaneous Questions from the entire book.
26.
27.
28.
4
8
4
8
4
4
8
8
4
4
4
4
4
8
4
4
4
8
4
12
4
4
8
4
4
8
4
4
4
4
120
4
4
120
4
4
4
4
4
12
4
4
8
4
4
4
120
4
4
4
4
4
4
4
4
4
4
4
4
4
4
8
4
4
8
4
4
4
4
8
4
4
4
4
4
4
4
4
8
4
8
4
4
4
4
4
4
4
4
4
8
8
4
4
8
4
4
4
8
4
4
12
4
4
4
4
4
8
12
4
8
4
4
4
4
4
8
4
4
4
4
4
4
8
8
8
4
8
4
8
4
4
4
8
4
4
4
2016
(Architecture)
4
8
4
4
4
Online
10/4/16
4
4
4
4
8
4
4
4
4
4
4
4
Online
9/4/16
4
4
8
4
4
4
4
8
4
4
JEE (Main)
(Offline) 2016
17.
18.
19.
20.
21.
22.
23.
24.
25.
8
Architecture
2017
8.
9.
10.
11.
12.
13.
14.
15.
16.
Sets, Relations and Functions
Complex Numbers
Quadratic Equations
Determinants
Matrices
Permutations and Combinations
Mathematical Induction and Binomial
Theorem
Progressions
Limits and Continuity
Differentiability and Differentiation
Applications of Derivatives
Indefinite Integration
Definite Integrals
Areas by Integration
Differential Equations
Cartesian System of Rectangular
Coordinates and Straight Lines
Circles and Systems of Circles
Parabola
Ellipse
Hyperbola
Three Dimensional Geometry
Vectors
Statistics
Probability
Trigonometrical Ratios, Identities
and Equations
Inverse Trigonometric Functions
Heights and Distances
Mathematical Reasoning
Total
Online
9/4/17
1.
2.
3.
4.
5.
6.
7.
Chapter Name
Online
8/4/17
Chap.
No.
JEE (Main)
(Offline) 2017
Trend Analysis
8
4
4
8
4
4
8
4
4
4
8
4
4
4
4
8
4
4
12
4
4
4
4
4
4
120
4
4
120
4
120
4
4
120
4
120
About JEE Main
1. Introduction and Scheme of Examination
The Joint Entrance Examination from the year 2013 for admission to the undergraduate programmes in Engineering is
being held in two parts, JEE-Main and JEE-Advanced. Only the top 2,00,000 candidates (including all categories) based
on performance in JEE Main will qualify to appear in the JEE Advanced examination. Admissions to IITs will be based
only on category-wise All India Rank (AIR) in JEE Advanced, subject to condition that such candidates are in the top
20 percentile categories. Admission to NITs will be based on 40% weightage for performance in Class XII board marks
(normalized) and the remainder 60% weightage would be given to performance in JEE Main and a combined All India
Rank (AIR) would be decided accordingly.
In case any State opts to admit students in the engineering Colleges affiliated to state Universities where States require
separate merit list to be provided based on relative weightages adopted by the states, then the merit list shall be prepared
with such relative weightages as may be indicated by States.
2. Eligibility Criteria and List of Qualifying Examinations for JEE(Main) Exam
The minimum academic qualification for appearing in JEE(Main) is that the candidate must have passed in final examination of 10+2 (Class XII) or its equivalent referred to as the qualifying examination (see below).
However, admission criteria in the concerned institution/university will be followed as prescribed by concerned
university/institution and as per the guidelines & criteria prescribed by AICTE.
Qualifying Examinations
List of Qualifying Examinations
(i) The +2 level examination in the 10+2 pattern of examination of any recognized Central/State Board of Secondary
Examination, such as Central Board of Secondary Education, New Delhi, and Council for Indian School Certificate Examination, New Delhi
(ii) Intermediate or two-year Pre-University Examination conducted by a recognized Board/University.
(iii) Final Examination of the two-year course of the Joint Services Wing of the National Defence Academy.
(iv) Any Public School/Board/University Examination in India or in foreign countries recognized by the Association
of Indian Universities as equivalent to 10+2 system.
(v) H.S.C. Vocational Examination.
(vi) A pass grade in the Senior Secondary School Examination conducted by the National Open School with a minimum of five subjects.
(vii) 3 or 4-year diploma recognized by AICTE or a State Board of Technical Education.
xii About JEE Main
3. Pattern of Examination
Subject combination for each paper and type of questions in each paper are given below:
Subjects
Type of Questions
Duration
Paper 1
Physics, Chemistry & Mathematics
Objective type questions with equal weightage
to Physics, Chemistry & Mathematics
3 Hours
Paper 2
Mathematics – Part I
Objective type questions
3 Hours
Aptitude Test – Part II &
Objective type questions
Drawing Test – Part III
Questions to test Drawing Aptitude
Requirement of papers for different courses is given in the table below:
Course
Papers
B.E/B.TECH
Paper – 1
B.ARCH/B. PLANNING
Paper – 2
Scoring and Negative Marking
There will be objective type questions with four options having single correct answer. For each incorrect response,
one fourth (1/4) of the total marks allotted to the question would be deducted. No deduction from the total score will,
however, be made if no response is indicated for an item in the answer sheet.
Syllabus
Unit 1: Sets, Relations and Functions
Sets and their representation; Union, intersection and complement of sets and their algebraic properties;
Power set; Relation, Types of relations, equivalence relations, functions;. one-one, into and onto functions,
composition of functions.
Unit 2: Complex Numbers and Quadratic Equations
Complex numbers as ordered pairs of reals, Representation of complex numbers in the form a + ib and their
representation in a plane, Argand diagram, algebra of complex numbers, modulus and argument (or amplitude)
of a complex number, square root of a complex number, triangle inequality, Quadratic equations in real
and complex number system and their solutions. Relation between roots and co-efficients, nature of roots,
formation of quadratic equations with given roots.
Unit 3: Matrices and Determinants
Matrices, algebra of matrices, types of matrices, determinants and matrices of order two and three. Properties
of determinants, evaluation of determinants, area of triangles using determinants. Adjoint and evaluation
of inverse of a square matrix using determinants and elementary transformations, Test of consistency and
solution of simultaneous linear equations in two or three variables using determinants and matrices.
Unit 4: Permutations and Combinations
Fundamental principle of counting, permutation as an arrangement and combination as selection, Meaning of
P (n, r) and C (n, r), simple applications.
Unit 5: Mathematical Induction
Principle of Mathematical Induction and its simple applications.
Unit 6: Binomial Theorem and its Simple Applications
Binomial theorem for a positive integral index, general term and middle term, properties of Binomial
coefficients and simple applications.
Unit 7: Sequences and Series
Arithmetic and Geometric progressions, insertion of arithmetic, geometric means between two given numbers.
Relation between A.M. and G.M. Sum upto n terms of special series: Ân, Ân2, Ân3. Arithmetico - Geometric
progression.
Unit 8: Limit, Continuity and Differentiability
Real—valued functions, algebra of functions, polynomials, rational, trigonometric, logarithmic and
exponential functions, inverse functions. Graphs of simple functions. Limits, continuity and differentiability.
Differentiation of the sum, difference, product and quotient of two functions. Differentiation of trigonometric,
inverse trigonometric, logarithmic, exponential, composite and implicit functions; derivatives of order upto
two. Rolle’s and Lagrange’s Mean Value Theorems. Applications of derivatives: Rate of change of quantities,
xiv Syllabus
monotonic—increasing and decreasing functions, Maxima and minima of functions of one variable, tangents
and normals.
Unit 9: Integral Calculus
Integral as an anti-derivative. Fundamental integrals involving algebraic, trigonometric, exponential and
logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using
trigonometric identities.
Evaluation of simple integrals of the type
dx
Ú x 2 ± a2 , Ú
Ú
Ú
dx
2
x ±a
dx
ax 2 + bx + c
,Ú
a2 ± x 2 dx and
2
,Ú
dx
2
a -x
2
( px + q ) dx
2
ax + bx + c
Ú
,Ú
,Ú
dx
2
a -x
2
,Ú
dx
2
ax + bx + c
( px + q ) dx
ax 2 + bx + c
x 2 - a2 dx
Integral as limit of a sum. Fundamental Theorem of Calculus. Properties of definite integrals. Evaluation of
definite integrals, determining areas of the regions bounded by simple curves in standard form.
Unit 10: Differential Equations
Ordinary differential equations, their order and degree. Formation of differential equations. Solution of
differential equations by the method of separation of variables, solution of homogeneous and linear differential
equations of the type:
dy
+ p( x ) y = q ( x )
dx
Unit 11: Co-ordinate Geometry
Cartesian system of rectangular co-ordinates in a plane, distance formula, section formula, locus and its equation,
translation of axes, slope of a line, parallel and perpendicular lines, intercepts of a line on the coordinate axes.
Straight lines
Various forms of equations of a line, intersection of lines, angles between two lines, conditions for concurrence
of three lines, distance of a point from a line, equations of internal and external bisectors of angles between two
lines, coordinates of centroid, orthocentre and circumcentre of a triangle, equation of family of lines passing
through the point of intersection of two lines.
Circles, conic sections
Standard form of equation of a circle, general form of the equation of a circle, its radius and centre, equation
of a circle when the end points of a diameter are given, points of intersection of a line and a circle with the
centre at the origin and condition for a line to be tangent to a circle, equation of the tangent. Sections of cones,
equations of conic sections (parabola, ellipse and hyperbola) in standard forms, condition for y = mx + c to
be a tangent and point (s) of tangency.
Unit 12: Three Dimensional Geometry
Coordinates of a point in space, distance between two points, section formula, direction ratios and direction
cosines, angle between two intersecting lines. Skew lines, the shortest distance between them and its equation.
Equations of a line and a plane in different forms, intersection of a line and a plane, coplanar lines.
Syllabus xv
Unit 13: Vector Algebra
Vectors and scalars, addition of vectors, components of a vector in two dimensions and three dimensional
space, scalar and vector products, scalar and vector triple product.
Unit 14: Statistics and Probability
Measures of Dispersion: Calculation of mean, median, mode of grouped and ungrouped data. Calculation of
standard deviation, variance and mean deviation for grouped and ungrouped data.
Probability: Probability of an event, addition and multiplication theorems of probability, Baye’s theorem,
probability distribution of a random variate, Bernoulli trials and Binomial distribution.
Unit 15: Trigonometry
Trigonometrical identities and equations. Trigonometrical functions. Inverse trigonometrical functions and
their properties. Heights and Distances.
Unit 16: Mathematical Reasoning
Statements, logical operations And, or, implies, implied by, if and only if. Understanding of tautology,
contradiction, converse and contrapositive.
Contents
To Our Readers…
Trend Analysis
About JEE Main
Syllabus
Format of Questions in this Book
1. Sets, Relations and Functions
v
ix
xi
xiii
xxix
1.1–1.40
Finite Set and Infinite Set 1.1
Algebra of Sets 1.1
Some Properties of Operations on Sets 1.2
Some Definitions 1.3
Algebraic Operations on Functions 1.4
Domains and Ranges of Some Functions 1.4
Types of Functions 1.4
Graphs of Some Functions 1.5
Inverse of a Function 1.9
Direct and Inverse Images 1.9
Solved Examples: Concept-based: Straight Objective Type Questions 1.10
LEVEL 1: Straight Objective Type Questions 1.12
Assertion-Reason Type Questions 1.19
LEVEL 2: Straight Objective Type Questions 1.20
Exercises: Concept-based: Straight Objective Type Questions 1.24
LEVEL 1: Straight Objective Type Questions 1.24
Assertion-Reason Type Questions 1.27
LEVEL 2: Straight Objective Type Questions 1.27
Previous Years’ AIEEE/JEE Main Questions 1.28
Previous Years’ B-Architecture Entrance Examination Questions 1.30
Answers 1.31
Hints and Solutions 1.32
2. Complex Numbers
Definitions 2.1
Algebraic Operations with Complex Numbers 2.1
Conjugate of Complex Number 2.1
Modulus of a Complex Number 2.2
Geometrical Representation of Complex Numbers 2.2
Argument of a Complex Number 2.2
Polar Form of a Complex Number 2.4
Vector Representation of Complex Numbers 2.4
Geometrical Representation of Algebraic Operations on Complex Numbers 2.4
2.1–2.59
xviii Contents
Some Important Geometrical Results and Equations 2.6
Recognizing Some Loci by Inspection 2.11
Geometric Interpretation of Multiplying a Complex Number by eia. 2.12
De Moivre’s Theorem and its Applications 2.13
Solved Examples: Concept-based: Straight Objective Type Questions 2.15
LEVEL 1: Straight Objective Type Questions 2.17
Assertion-Reason Type Questions 2.27
LEVEL 2: Straight Objective Type Questions 2.30
Exercises: Concept-based: Straight Objective Type Questions 2.34
LEVEL 1: Straight Objective Type Questions 2.35
Assertion-Reason Type Questions 2.39
LEVEL 2: Straight Objective Type Questions 2.39
Previous Years’ AIEEE/JEE Main Questions 2.41
Previous Years’ B-Architecture Entrance Examination Questions 2.43
Answers 2.44
Hints and Solutions 2.45
3. Quadratic Equations
3.1–3.58
Quadratic Equations 3.1
Nature of Roots 3.1
Relation between Roots and Coefficients 3.1
Quadratic Expression and its Graph 3.1
Position of Roots of a quadratic Equation 3.2
Conditions for a Number k to Lie Between the Roots of a Quadratic Equation 3.3
Conditions for Exactly one Root of a Quadratic Equation to Lie in the Interval (k1, k2) where k1 < k2 3.3
Conditions for Both the Roots of a Quadratic Equation to Lie in the Interval (k1, k2) where k1 < k2. 3.4
Conditions for a Quadratic Equation to Have a Repeated Root 3.5
Condition for two Quadratic Equations to Have a Common Root 3.5
Condition for Two Quadratic Equations to Have the Same Roots 3.5
Equations of Higher Degree 3.5
Cubic and Biquadratic Equations 3.5
Transformation of Equations 3.6
Descartes Rule of Signs for the Roots of a Polynomial 3.6
Some Hints for Solving Polynomial Equations 3.6
Use of Continuity and Differentiability to Find Roots of a Polynomial Equation 3.7
Solved Examples: Concept-based: Straight Objective Type Questions 3.7
LEVEL 1: Straight Objective Type Questions 3.10
Assertion-Reason Type Questions 3.24
LEVEL 2: Straight Objective Type Questions 3.27
Exercises: Concept-based: Straight Objective Type Questions 3.31
LEVEL 1: Straight Objective Type Questions 3.32
Assertion-Reason Type Questions 3.37
LEVEL 2: Straight Objective Type Questions 3.38
Previous Years’ AIEEE/JEE Main Questions 3.39
Previous Years’ B-Architecture Entrance Examination Questions 3.42
Answers 3.42
Hints and Solutions 3.43
4. Determinants
Evaluation of Determinants 4.1
4.1–4.60
Contents xix
Minors and Cofactors 4.1
Properties of Determinants 4.2
Some Tips for Quick Evaluation of Determinants 4.3
Linear Equations 4.4
Cramer’s Rule 4.4
Solved Examples: Concept-based: Straight Objective Type Questions 4.4
LEVEL 1: Straight Objective Type Questions 4.8
Assertion-Reason Type Questions 4.24
LEVEL 2: Straight Objective Type Questions 4.26
Exercises: Concept-based: Straight Objective Type Questions 4.31
LEVEL 1: Straight Objective Type Questions 4.33
Assertion-Reason Type Questions 4.38
LEVEL 2: Straight Objective Type Questions 4.39
Previous Years’ AIEEE/JEE Main Questions 4.40
Previous Years’ B-Architecture Entrance Examination Questions 4.42
Answers 4.43
Hints and Solutions 4.43
5. Matrices
5.1–5.46
The Algebra of Matrices 5.1
Transpose of a Matrix 5.2
Adjoint and Inverse of a Matrix 5.2
Some Definitions and Results 5.3
Rank of a Matrix 5.3
System of Linear Equations 5.5
Solution of a System of Equation AX = B when A is a square matrix 5.6
Solution of a System of Homogeneous Linear Equations Ax = 0 5.6
Solved Examples: Concept-based: Straight Objective Type Questions 5.7
LEVEL 1: Straight Objective Type Questions 5.9
Assertion-Reason Type Questions 5.20
LEVEL 2: Straight Objective Type Questions 5.22
Exercises: Concept-based: Straight Objective Type Questions 5.24
LEVEL 1: Straight Objective Type Questions 5.25
Assertion-Reason Type Questions 5.28
LEVEL 2: Straight Objective Type Questions 5.28
Previous Years’ AIEEE/JEE Main Questions 5.30
Previous Years’ B-Architecture Entrance Examination Questions 5.33
Answers 5.33
Hints and Solutions 5.34
6. Permutations and Combinations
Fundamental Principles of Counting 6.1
Permutations and Combinations 6.1
Some Important Results 6.2
Circular Permutation 6.2
Some Important Identities 6.2
Division of Identical Objects 6.2
Multiplication of Two Infinite Series 6.3
Distribution into Groups 6.3
Selection 6.4
6.1–6.40
xx Contents
Exponent of Prime p in n! 6.4
Solved Examples: Concept-based: Straight Objective Type Questions 6.4
LEVEL 1: Straight Objective Type Questions 6.6
Assertion-Reason Type Questions 6.17
LEVEL 2: Straight Objective Type Questions 6.19
Exercises: Concept-based: Straight Objective Type Questions 6.21
LEVEL 1: Straight Objective Type Questions 6.22
Assertion-Reason Type Questions 6.25
LEVEL 2: Straight Objective Type Questions 6.26
Previous Years’ AIEEE/JEE Main Questions 6.26
Previous Years’ B-Architecture Entrance Examination Questions 6.29
Answers 6.29
Hints and Solutions 6.30
7. Mathematical Induction and Binomial Theorem
7.1–7.51
Principle of Mathematical Induction 7.1
Binomial Theorem (for a positive integral index) 7.1
Properties of the Binomial Expansion 7.1
Middle Term(s) 7.1
Some Properties of the Binomial Coefficients 7.2
Reversing Technique 7.2
Multinomial Theorem 7.3
Some useful tips and tricks 7.3
Some Particular Expansions 7.3
Solved Examples: Concept-based: Straight Objective Type Questions 7.3
LEVEL 1: Straight Objective Type Questions 7.6
Assertion-Reason Type Questions 7.20
LEVEL 2: Straight Objective Type Questions 7.22
Exercises: Concept-based: Straight Objective Type Questions 7.28
LEVEL 1: Straight Objective Type Questions 7.28
Assertion-Reason Type Questions 7.31
LEVEL 2: Straight Objective Type Questions 7.32
Previous Years’ AIEEE/JEE Main Questions 7.33
Previous Years’ B-Architecture Entrance Examination Questions 7.35
Answers 7.36
Hints and Solutions 7.36
8. Progressions
Sequences 8.1
Arithmetic Progression 8.1
Geometric Progression 8.2
Arithmetico-Geometric Sequence 8.2
Harmonic Progression 8.3
Summation of Some Series of Natural Numbers 8.3
Sum of the Products of Two Terms of a Sequence 8.3
Method of Difference for Summation of Series 8.4
Solved Examples: Concept-based: Straight Objective Type Questions 8.4
LEVEL 1: Straight Objective Type Questions 8.6
Assertion-Reason Type Questions 8.22
LEVEL 2: Straight Objective Type Questions 8.24
8.1–8.55
Contents xxi
Exercises: Concept-based: Straight Objective Type Questions 8.27
LEVEL 1: Straight Objective Type Questions 8.28
Assertion-Reason Type Questions 8.32
LEVEL 2: Straight Objective Type Questions 8.32
Previous Years’ AIEEE/JEE Main Questions 8.33
Previous Years’ B-Architecture Entrance Examination Questions 8.37
Answers 8.38
Hints and Solutions 8.38
9. Limits and Continuity
9.1–9.43
Limits 9.1
One-sided Limits 9.1
Frequently Used Limits 9.2
Some Theorems on Limits 9.2
Computation of Limits 9.2
L’Hôpital’s Rules for Calculating Limits 9.3
Use of Series Expansion in Finding Limits 9.4
Sequential Limits 9.4
Some Useful Results on Limits 9.5
Continuity 9.5
Functions Continuous on a Closed Interval 9.6
Solved Examples: Concept-based: Straight Objective Type Questions 9.6
LEVEL 1: Straight Objective Type Questions 9.8
Assertion-Reason Type Questions 9.16
LEVEL 2: Straight Objective Type Questions 9.18
Exercises: Concept-based: Straight Objective Type Questions 9.24
LEVEL 1: Straight Objective Type Questions 9.25
Assertion-Reason Type Questions 9.27
LEVEL 2: Straight Objective Type Questions 9.28
Previous Years’ AIEEE/JEE Main Questions 9.30
Previous Years’ B-Architecture Entrance Examination Questions 9.32
Answers 9.33
Hints and Solutions 9.33
10. Differentiability and Differentiation
Differentiability 10.1
Differentiability on An Interval 10.2
Higher Order Differentiation 10.2
Some Formulae of Differentiation 10.2
Differentiation of Some Elementary Functions 10.2
Differentiation of Implicit Functions 10.3
Differentiation of Functions Represented Parametrically 10.3
Logarithmic Differentiation 10.3
Leibnitz Theorem and nth Derivatives 10.4
Solved Examples: Concept-based: Straight Objective Type Questions 10.4
LEVEL 1: Straight Objective Type Questions 10.6
Assertion-Reason Type Questions 10.15
LEVEL 2: Straight Objective Type Questions 10.16
Exercises: Concept-based: Straight Objective Type Questions 10.22
LEVEL 1: Straight Objective Type Questions 10.23
10.1–10.46
xxii Contents
Assertion-Reason Type Questions 10.26
LEVEL 2: Straight Objective Type Questions 10.26
Previous Years’ AIEEE/JEE Main Questions 10.28
Previous Years’ B-Architecture Entrance Examination Questions 10.31
Answers 10.32
Hints and Solutions 10.32
11. Applications of Derivatives
11.1–11.51
The Derivative as a Rate of Change 11.1
Tangent and Normal 11.2
Angle Between Two Curves 11.3
The Rolle’s and Lagrange’s Theorems 11.3
Monotonicity 11.3
Maxima and Minima 11.4
Solved Examples: Concept-based: Straight Objective Type Questions 11.5
LEVEL 1: Straight Objective Type Questions 11.7
Assertion-Reason Type Questions 11.16
LEVEL 2: Straight Objective Type Questions 11.18
Exercises: Concept-based: Straight Objective Type Questions 11.23
LEVEL 1: Straight Objective Type Questions 11.23
Assertion-Reason Type Questions 11.26
LEVEL 2: Straight Objective Type Questions 11.26
Previous Years’ AIEEE/JEE Main Questions 11.28
Previous Years’ B-Architecture Entrance Examination Questions 11.32
Answers 11.33
Hints and Solutions 11.34
12. Indefinite Integration
12.1–12.51
Table of Basic Formula 12.1
A List of Evaluation Techniques 12.2
Some Tips for Simplifying Computations 12.7
Some Tips for Partial Fractions 12.7
Solved Examples: Concept-based: Straight Objective Type Questions 12.9
LEVEL 1: Straight Objective Type Questions 12.11
Assertion-Reason Type Questions 12.21
LEVEL 2: Straight Objective Type Questions 12.23
Exercises: Concept-based: Straight Objective Type Questions 12.29
LEVEL 1: Straight Objective Type Questions 12.30
Assertion-Reason Type Questions 12.33
LEVEL 2: Straight Objective Type Questions 12.33
Previous Years’ AIEEE/JEE Main Questions 12.36
Previous Years’ B-Architecture Entrance Examination Questions 12.38
Answers 12.39
Hints and Solutions 12.40
13. Definite Integrals
The Newton-Leibnitz Formula 13.1
Definite Integral as the Limit of a Sum 13.1
Properties of Definite Integrals 13.1
13.1–13.54
Contents xxiii
Integrals with Infinite Limits 13.2
Reduction Formulae for
p /2
Ú0
sin n x d x and Ú
p /2
0
sin p x cosq xd x
13.3
Solved Examples: Concept-based: Straight Objective Type Questions 13.3
LEVEL 1: Straight Objective Type Questions 13.5
Assertion-Reason Type Questions 13.17
LEVEL 2: Straight Objective Type Questions 13.18
Exercises: Concept-based: Straight Objective Type Questions 13.25
LEVEL 1: Straight Objective Type Questions 13.26
Assertion-Reason Type Questions 13.29
LEVEL 2: Straight Objective Type Questions 13.29
Previous Years’ AIEEE/JEE Main Questions 13.32
Previous Years’ B-Architecture Entrance Examination Questions 13.35
Answers 13.37
Hints and Solutions 13.37
14. Areas by Integration
14.1–14.25
Solved Examples: Concept-based: Straight Objective Type Questions 14.4
LEVEL 1: Straight Objective Type Questions 14.5
Assertion-Reason Type Questions 14.9
LEVEL 2: Straight Objective Type Questions 14.9
Exercises: Concept-based: Straight Objective Type Questions 14.12
LEVEL 1: Straight Objective Type Questions 14.12
Assertion-Reason Type Questions 14.13
Previous Years’ AIEEE/JEE Main Questions 14.14
Previous Years’ B-Architecture Entrance Examination Questions 14.15
Answers 14.15
Hints and Solutions 14.16
15. Differential Equations
Formation of a Differential Equation 15.1
Methods of Solving Differential Equations Solution or Integral of a Differential Equation 15.2
Equations of First Order and First Degree 15.2
Orthogonal Trajectory 15.3
Differential Equations of First Order but not of First Degree 15.3
Second Order but of Degree One 15.4
Solved Examples: Concept-based: Straight Objective Type Questions 15.4
LEVEL 1: Straight Objective Type Questions 15.7
Assertion-Reason Type Questions 15.14
LEVEL 2: Straight Objective Type Questions 15.15
Exercises: Concept-based: Straight Objective Type Questions 15.19
LEVEL 1: Straight Objective Type Questions 15.20
Assertion-Reason Type Questions 15.22
LEVEL 2: Straight Objective Type Questions 15.22
Previous Years’ AIEEE/JEE Main Questions 15.24
Previous Years’ B-Architecture Entrance Examination Questions 15.26
Answers 15.27
Hints and Solutions 15.27
15.1–15.40
xxiv Contents
16. Cartesian System of Rectangular Coordinates and Straight Lines
16.1–16.56
Results Regarding Points in a Plane 16.1
Standard Forms of the Equation of a Line 16.2
Some Results for Two or More Lines 16.3
Some Useful Points 16.4
Locus of a Point 16.4
Change of Axes 16.5
Equation of Family of Lines 16.5
Solved Examples: Concept-based: Straight Objective Type Questions 16.6
LEVEL 1: Straight Objective Type Questions 16.9
Assertion-Reason Type Questions 16.21
LEVEL 2: Straight Objective Type Questions 16.23
Exercises: Concept-based: Straight Objective Type Questions 16.29
LEVEL 1: Straight Objective Type Questions 16.30
Assertion-Reason Type Questions 16.34
LEVEL 2: Straight Objective Type Questions 16.34
Previous Years’ AIEEE/JEE Main Questions 16.36
Previous Years’ B-Architecture Entrance Examination Questions 16.40
Answers 16.41
Hints and Solutions 16.42
17. Circles and Systems of Circles
17.1–17.56
Definition of a Circle 17.1
Equations of a Circle 17.1
Some Results Regarding Circles 17.2
Special Forms of Equation of a Circle 17.3
Systems of Circles 17.3
Common Tangents to Two Circles 17.4
Solved Examples: Concept-based: Straight Objective Type Questions 17.5
LEVEL 1: Straight Objective Type Questions 17.8
Assertion-Reason Type Questions 17.19
LEVEL 2: Straight Objective Type Questions 17.20
Exercises: Concept-based: Straight Objective Type Questions 17.29
LEVEL 1: Straight Objective Type Questions 17.31
Assertion-Reason Type Questions 17.34
LEVEL 2: Straight Objective Type Questions 17.34
Previous Years’ AIEEE/JEE Main Questions 17.37
Previous Years’ B-Architecture Entrance Examination Questions 17.39
Answers 17.40
Hints and Solutions 17.41
18. Parabola
Conic Section 18.1
Parabola 18.1
Solved Examples: Concept-based: Straight Objective Type Questions 18.5
LEVEL 1: Straight Objective Type Questions 18.8
Assertion-Reason Type Questions 18.13
LEVEL 2: Straight Objective Type Questions 18.15
Exercises: Concept-based: Straight Objective Type Questions 18.18
18.1–18.37
Contents xxv
LEVEL 1: Straight Objective Type Questions 18.19
Assertion-Reason Type Questions 18.21
LEVEL 2: Straight Objective Type Questions 18.22
Previous Years’ AIEEE/JEE Main Questions 18.23
Previous Years’ B-Architecture Entrance Examination Questions 18.24
Answers 18.25
Hints and Solutions 18.26
19. Ellipse
19.1–19.37
Definition 1 19.1
Definition 2 19.1
Solved Examples: Concept-based: Straight Objective Type Questions 19.5
LEVEL 1: Straight Objective Type Questions 19.8
Assertion-Reason Type Questions 19.14
LEVEL 2: Straight Objective Type Questions 19.17
Exercises: Concept-based: Straight Objective Type Questions 19.20
LEVEL 1: Straight Objective Type Questions 19.21
Assertion-Reason Type Questions 19.23
LEVEL 2: Straight Objective Type Questions 19.24
Previous Years’ AIEEE/JEE Main Questions 19.25
Previous Years’ B-Architecture Entrance Examination Questions 19.26
Answers 19.27
Hints and Solutions 19.27
20. Hyperbola
20.1–20.35
Solved Examples: Concept-based: Straight Objective Type Questions 20.4
LEVEL 1: Straight Objective Type Questions 20.7
Assertion-Reason Type Questions 20.13
LEVEL 2: Straight Objective Type Questions 20.16
Exercises: Concept-based: Straight Objective Type Questions 20.18
LEVEL 1: Straight Objective Type Questions 20.20
Assertion-Reason Type Questions 20.21
LEVEL 2: Straight Objective Type Questions 20.22
Previous Years’ AIEEE/JEE Main Questions 20.23
Previous Years’ B-Architecture Entrance Examination Questions 20.25
Answers 20.25
Hints and Solutions 20.25
21. Three Dimensional Geometry
Coordinates 21.1
The Plane 21.2
The Straight Line 21.4
Vectorial Equations 21.6
Solved Examples: Concept-based: Straight Objective Type Questions 21.7
LEVEL 1: Straight Objective Type Questions 21.10
Assertion-Reason Type Questions 21.19
LEVEL 2: Straight Objective Type Questions 21.20
Exercises: Concept-based: Straight Objective Type Questions 21.27
21.1–21.56
xxvi Contents
LEVEL 1: Straight Objective Type Questions 21.28
Assertion-Reason Type Questions 21.31
LEVEL 2: Straight Objective Type Questions 21.32
Previous Years’ AIEEE/JEE Main Questions 21.33
Previous Years’ B-Architecture Entrance Examination Questions 21.38
Answers 21.39
Hints and Solutions 21.40
22. Vectors
22.1–22.39
Linear Combinations 22.1
Scalar or Dot Product 22.2
Vector or Cross Product 22.2
Scalar Triple Product 22.2
Vector Triple Product 22.3
Geometrical and Physical Applications 22.3
Solved Examples: Concept-based: Straight Objective Type Questions 22.3
LEVEL 1: Straight Objective Type Questions 22.5
Assertion-Reason Type Questions 22.12
LEVEL 2: Straight Objective Type Questions 22.13
Exercises: Concept-based: Straight Objective Type Questions 22.17
LEVEL 1: Straight Objective Type Questions 22.18
Assertion-Reason Type Questions 22.20
LEVEL 2: Straight Objective Type Questions 22.20
Previous Years’ AIEEE/JEE Main Questions 22.22
Previous Years’ B-Architecture Entrance Examination Questions 22.25
Answers 22.26
Hints and Solutions 22.27
23. Statistics
Measures of Central Tendency 23.1
Arithmetic Mean 23.1
Properties of Arithmetic Mean 23.1
Geometric Mean 23.2
Harmonic Mean 23.2
Median 23.2
Mode 23.2
Dispersion 23.3
Coefficient of Dispersion (C.D.) 23.3
Solved Examples: Concept-based: Straight Objective Type Questions 23.4
LEVEL 1: Straight Objective Type Questions 23.5
Assertion-Reason Type Questions 23.8
LEVEL 2: Straight Objective Type Questions 23.9
Exercises: Concept-based: Straight Objective Type Questions 23.10
LEVEL 1: Straight Objective Type Questions 23.11
Assertion-Reason Type Questions 23.12
LEVEL 2: Straight Objective Type Questions 23.12
Previous Years’ AIEEE/JEE Main Questions 23.13
Previous Years’ B-Architecture Entrance Examination Questions 23.15
Answers 23.15
Hints and Solutions 23.16
23.1–23.20
Contents xxvii
24. Probability
24.1–24.50
Definitions 24.1
Classical Definition of Probability 24.1
Notations 24.1
A Few Theorems on Probability 24.1
Conditional Probability 24.2
Multiplication Theorem 24.2
Total Probability Theorem 24.3
Bayes’ Rule 24.3
Theorem (Bayes’ Rule) 24.3
Independent Events 24.3
Random Variables 24.3
Binomial Distributions 24.4
Solved Examples: Concept-based: Straight Objective Type Questions 24.5
LEVEL 1: Straight Objective Type Questions 24.7
Assertion-Reason Type Questions 24.22
LEVEL 2: Straight Objective Type Questions 24.24
Exercises: Concept-based: Straight Objective Type Questions 24.27
LEVEL 1: Straight Objective Type Questions 24.28
Assertion-Reason Type Questions 24.33
LEVEL 2:Straight Objective Type Questions 24.33
Previous Years’ AIEEE/JEE Main Questions 24.34
Previous Years’ B-Architecture Entrance Examination Questions 24.37
Answers 24.38
Hints and Solutions 24.39
25. Trigonometrical Ratios, Identities and Equations
Introduction 25.1
Domain and Range of Trigonometrical Functions 25.1
Some Basic Formulae 25.1
Allied or Related Angles 25.2
Compound Angles 25.2
Some Important Results 25.3
Identities 25.3
Conditional Identities 25.3
General Solutions of Trigonometrical Equations 25.4
Solved Examples: Concept-based: Straight Objective Type Questions 25.4
LEVEL 1: Straight Objective Type Questions 25.8
Assertion-Reason Type Questions 25.16
LEVEL 2: Straight Objective Type Questions 25.17
Exercises: Concept-based: Straight Objective Type Questions 25.23
LEVEL 1: Straight Objective Type Questions 25.25
Assertion-Reason Type Questions 25.27
LEVEL 2: Straight Objective Type Questions 25.28
Previous Years’ AIEEE/JEE Main Questions 25.30
Previous Years’ B-Architecture Entrance Examination Questions 25.32
Answers 25.32
Hints and Solutions 25.33
25.1–25.47
xxviii Contents
26. Inverse Trigonometric Functions
26.1–26.25
Inverse Trigonometric Functions and Formulae 26.1
Solved Examples: Concept-based: Straight Objective Type Questions 26.5
LEVEL 1: Straight Objective Type Questions 26.6
Assertion-Reason Type Questions 26.11
LEVEL 2: Straight Objective Type Questions 26.12
Exercises: Concept-based: Straight Objective Type Questions 26.14
LEVEL 1: Straight Objective Type Questions 26.14
Assertion-Reason Type Questions 26.16
LEVEL 2: Straight Objective Type Questions 26.16
Previous Years’ AIEEE/JEE Main Questions 26.17
Previous Years’ B-Architecture Entrance Examination Questions 26.18
Answers 26.18
Hints and Solutions 26.19
27. Heights and Distances
27.1–27.37
Angles of Elevation and Depression 27.1
Solved Examples: Concept-based: Straight Objective Type Questions 27.2
LEVEL 1: Straight Objective Type Questions 27.3
Assertion-Reason Type Questions 27.13
LEVEL 2: Straight Objective Type Questions 27.14
Exercises: Concept-based: Straight Objective Type Questions 27.20
LEVEL 1: Straight Objective Type Questions 27.21
Assertion-Reason Type Questions 27.23
LEVEL 2: Straight Objective Type Questions 27.23
Previous Years’ AIEEE/JEE Main Questions 27.25
Previous Years’ B-Architecture Entrance Examination Questions 27.26
Answers 27.26
Hints and Solutions 27.26
28. Mathematical Reasoning
28.1–28.16
Some Definitions 28.1
Solved Examples: Concept-based: Straight Objective Type Questions 28.2
LEVEL 1: Straight Objective Type Questions 28.3
Assertion-Reason Type Questions 28.6
LEVEL 2: Straight Objective Type Questions 28.7
Exercises: Concept-based: Straight Objective Type Questions 28.8
LEVEL 1: Straight Objective Type Questions 28.8
Assertion-Reason Type Questions 28.9
LEVEL 2: Straight Objective Type Questions 28.9
Previous Years’ AIEEE/JEE Main Questions 28.10
Previous Years’ B-Architecture Entrance Examination Questions 28.11
Answers 28.12
Hints and Solutions 28.12
Architecture Entrance – 2017
JEE (Main) 2017 Questions with Solutions Mathematics (8th April - online)
JEE (Main) 2017 Questions with Solution (9th April – online)
JEE Main 2017 (Offline)
1–8
9–16
17–24
25–32
Format of Questions in this Book xxix
Format of Questions in this Book
Straight Objective Type Questions
Each Question has four choices—(a), (b), (c) and (d) out of which ONLY ONE is correct
Assertion-Reason Type Questions
Each Question has four choices—(a), (b), (c) and (d) out of which ONLY ONE is correct
Write (a), (b), (c) or (d) according to the following rules.
(a) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
(b) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
(c) STATEMENT-1 is True, STATEMENT-2 is False
(d) STATEMENT-1 is False, STATEMENT-2 is True
CHAPTER ONE
By a set we mean any collection of objects. For example,
we may speak of the set of all living Indians, the set of all
letters of the English alphabet or the set of real numbers
less than 5. The objects constituting a set are elements or
members of the set.
If X is a set and an element x is a member of, or belongs
to the set X, then this is expressed, as x ΠX.
There are two methods of representing a set.
(a) Roster method (or tabular method) In this method, a set is represented by listing all its elements in
curly brackets and separating them by commas.
For example, set having elements 1, 2, 3, 5 only is
written as {1, 2, 3, 5}.
(b) Property method In this method a set is represented by stating all the properties which are satisfied
by the elements of the set and not by other elements
outside the set.
If X contains all values of ‘x’ for which the condition
P(x) is true, then we write X = {x : P(x)}.
Illustration
1
The set of all real roots of the equation x4 – 2x2 – 3 = 0 is
denoted by
{x : x ΠR, x4 Р2x2 Р3 = 0} or equivalently by
Illustration
{
}
3, - 3 .
2
The set of integers strictly between 2 and 8 is represented
by
{x : x ΠI, 2 < x < 8} or equivalently by {3, 4, 5, 6, 7}.
Illustration
3
The set of Presidents of India can be represented by
{x : x is/was President of India} or equivalently
A = {Rajendra Prasad, Sarvepalli Radhakrishnan,
Zakir Hussain, V.V. Giri, …….}
The number of elements of A or the cardinality of A is
denoted by n(A). In Illustration 3, n(A) = 13.
FINITE SET AND INFINITE SET
A set is called a finite set if it contains only finite number
of elements. A set which does not contain finite number of
elements is called infinite set.
Null set A set containing no element is called a null set
or empty set or void set and is denoted by ‘f’. Equivalently
P(x) is a property satisfied by no object at all.
Singleton set A set containing exactly one element is
called a singleton set.
If a set X has ‘r’ elements then we write n(X) (number of
elements in X) = r.
Subset A set ‘A’ is said to be subset of the set X if every
element of A is an element of X and we write A Õ X. A is
said to be a proper subset of X if A Õ X. and A π X, this
will be written as A Ã X . The denial of A Ã B is written as
π
A Ã/ B.
For two sets A and B, A = B if and only if A Õ B and
B Õ A.
The set of all subsets of X is called Power set of X denoted
by P(X) i.e. P(X) = {A : A Õ X}.
Some Basic Properties:
(i)
(i)
(iii)
(iv)
(v)
(vi)
AÕA
A Õ B, B Õ C then A Õ C
The only subset of f is f itself.
The subsets of {x} are f and {x}.
The subsets of {x, y} are f, {x}, {y}, {x, y}
If n(X) = r then number of all subsets of X will be
2r i.e. n(P(X)) = 2r.
ALGEBRA OF SETS
Union of two Sets
Union of two sets A and B is denoted by A » B
Complete Mathematics—JEE Main
1.2
A » B = {x : x Œ A
or x ΠB}
Clearly A Õ A » B,
BÕA»B
A
and,
«
B
A
Ê n ˆ n
4. A ~ Á « Ai ˜ = » ( A ~ Ai ).
Ë i =1 ¯ i =1
B
Ê n ˆ n
5. A ~ Á » Ai ˜ = « ( A ~ Ai ).
Ë i =1 ¯ i =1
A » B = B » A,
A » A = A,
Fig. 1.1
A » B = B if and only if A Õ B,
(A » B) » C = A » (B » C) (Associativity).
and
Intersection of Two Sets
Intersection of two sets A
and B is denoted by
A « B and, A « B = {x : x
ΠA and x ΠB}
Trivially, A « B Õ A,
A « B Õ B and A « B =
B « A, (A « B) « C =
A « (B « C) (Associativity).
A
B
A«B
Fig. 1.2
Complement of a Set
Complement of a set B in a set A is written as
A ~ B = {x : x Œ A, x œ B}
e.g.,
I ~ N = º – 3, – 2, – 1, 0}. The set of irrational
numbers = R ~ Q
Where I is the set of integers
N is the set of natural numbers
R is the set of real number and
Q is the set of rational numbers
In any discussion involving sets and their operations, we
presume that all these sets are subsets of a parent set called
Universal set, U. The complement of A in U is denoted by
Ac or A¢ = {x : x œ A}.
A
A « Ac = f
A » Ac = X
fc = X
Xc = f
(Ac )c = A
A ~ B = A « Bc
A Õ B if and only if Bc Õ Ac.
A ~ A = f.
(A ~ B) ~ C = (A ~ C) ~ B
A ~ (B » C) = (A ~ B) « (A ~ C)
If A and B are finite sets, then
n(A » B) = n(A) + n(B) – n(A « B)
17. If A, B and C are finite sets then
n(A » B » C) = n(A) + n(B) + n(C) – n(A « B)
– n(B « C) – n(A « C) + n(A « B « C)
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
Symmetric Difference of Two Sets
A D B= (A ~ B) » (B ~ A)
A
B
Fig. 1.4
A D A = f, A D B = B D A
Cartesian Product
B
A¢
A
Fig. 1.3
SOME PROPERTIES OF OPERATIONS
ON SETS
1. A « (B1 » B2 » ... » Bn) = (A « B1) » (A « B2)
» … » (A « Bn) (Intersection is distributive over
union)
2. A » (B1 « B2 « … « Bn) = (A » B1) « (A » B2) « …
« (A » Bn) (Union is distributive over intersection)
3. (A « B)c = Ac » Bc and (A » B)c = Ac « Bc.
(De Morgan’s Laws)
The cartesian product of sets A and B is denoted by A ¥ B
and, A ¥ B = {(a, b): a Œ A, b Œ B}
In general A ¥ B π B ¥ A, moreover,
A ¥ (B » C) = (A ¥ B) » (A ¥ C)
and
A ¥ (B « C) = (A ¥ B) « (A ¥ C)
Illustration
4
If A = {0, 1, 2, 3, 5}, B = {0, 1, 2, 3), C = {0, 1, 4, 5} then
B Õ A, C Õ/ A, A « C = {0, 1, 5}, B « C = {0, 1}, A ~ B =
{5}, A ~ C = {2, 3}, B ~ C = {2, 3}, C ~ B = {4, 5}, A D B
= {5}, A D C = {2, 3, 4} and B D C = {2, 3, 4, 5}
Illustration
5
Simplify (A » B » C ) « (A « Bc « C c )c « C c
Sets, Relations and Functions 1.3
(A » B » C ) « (A « Bc « C c )c « C c
c
Equivalence Relation
= [(A » B » C ) « (A » B » C )] « C
c
A relation R on a set X (i.e. R Ã X ¥ X) is said to be
= [(A « Ac) » (B » C] « C c
(i) reflexive if x R x for all x ŒX.
(ii) symmetric if x R y fi y R x, where x, y ΠX
(iii) transitive if x R y, y R z fi x R z, where x, y, z ΠX
= [f » (B » C)] « C c
= (B » C) « C c = (B « C c ) » (C « C c ).
= (B « C c) » f = B « C c = B ~ C.
Further, a relation R in a set X is said to be an equivalence
relation if it is reflexive, symmetric and transitive. R is
called anti-symmetric if x R y and y R x fi x = y.
Relations
A relation from a set of X to a
set Y is a subset R of X ¥ Y. If
(a, b) ΠR, we say a is related
to b and often write it as a R b.
If X = Y, we say that R is relation on X.
Let R be a relation from X to
Y. For each subset A of X, we
write
Y
6
Illustration
R
For x, y ΠI, write x R y if x Рy divisible by 6; this is an
equivalence relation and is usually written as x ∫ y (mod. 6).
X
Fig. 1.5
7
Illustration
The relation R on R defined as R = {(a, b): a £ b} is not
symmetric as (1, 3) Œ R but (3, 1) œ R but is transitive and
reflexive.
R(A) = {y ΠY: x R y for some x ΠA}
and call it the (direct) image of A under R, and for each
subset B of Y, we write R–1 (B) = {x Œ X : x R y for some y
ΠB} and call it the inverse image of B under R.
Y
B
8
Illustration
If A = {1, 2, 3}; let R = {1, 1), (2, 2) (3, 3), (1, 2) (2, 1)} then
R is reflexive, symmetric and transitive.
The number of equivalence relations that can be defined
on a set containing k elements is given
k -1
Bk =
 ( k n-1 )Bn
n=0
E.g.
B0 = 1, B1 = 1, B2 = 2, B3 =
( 20 ) B0 + ( 12 ) B1 + ( 22 ) B2
=1+2+2=5
R(A)
Functions
R-1(B)
A
X
Fig. 1.6
One can think of R–1 as a relation from Y to X, where
yR –1 x ¤ xRy
that is,
R –1 = {(y, x) : (x, y) Œ R}
R–1 is called the reverse of the relation R.
Domain of a relation R from a set X to a set Y is the set
of all first components of the elements of R, i.e., dom R =
{a ΠX: (a, b) ΠR for some b ΠR}
Let R be a relation from X to Y and S from Y to Z, then
SoR = {(x, z)| x ΠX, z ΠZ and $ y ΠY such that and (x, y)
ΠR and (y, z) ΠS.
If n(A), n(B) < •, then the number of relation from A to
B will be 2n(A) n(B)
A relation F from A to B is said to be a function if for each
a ΠA there exists a unique b ΠB such that (a, b) ΠF. b is
called image of ‘a’ under F and is denoted by F(a). Thus
every function is a relation but the converse may not be true.
If n(A) = n and n(B) = m then total number of functions
from A to B is nm.
SOME DEFINITIONS
Let A and B be two non-empty sets. A function f from A to B
can also be defined as a rule that assigns to each element in
the set A, one and only one element of the set B. In general,
the sets A and B need not be sets of real numbers. However,
we consider only those functions for which A and B are both
subsets of the real numbers. We shall denote by R, the set
of all real numbers.
The set A in the above definition is called the domain
of the function f. We usually denote it by dom f. If x is an
element in the domain of a function f, then the element that
1.4
Complete Mathematics—JEE Main
f associates with x is denoted by the symbol f (x), and is
called the image of x under f, or the value of f at x. The set
of all possible values of f (x) as x varies over the domain is
called the range of f. If f : A Æ B, then the range of f is a
subset of B and the set B is called co-domain of f.
Remark
If x is an element in the domain of a function f
a function requires that f assigns one and only one value to x.
This means that a function cannot be multiple-valued. For
function
example, the expression ± x
of x, since it assigns two values to each positive x.
ALGEBRAIC OPERATIONS ON
FUNCTIONS
1. If f and g are two functions, then sum of the functions
f + g, is defined for all x Œ dom f « dom g by
( f + g) (x) = f (x) + g(x).
2. If k is any real number and f is a function, then k f is
defined for all x Πdom f by (k f ) (x) = k f (x).
3. If f and g are two functions, then the pointwise product f g is defined for all x Œ dom f « dom g by
( f g) (x) = f (x) g(x).
4. If f and g are functions, then f/g is defined for all
x Œ dom (f) « dom(g) « {x: g(x) π 0} by (f/g)(x) =
f(x)/g(x).
5. Composition of functions Let f : A Æ B and g : B Æ C
be functions, then gof : A Æ C defined by (gof ) (x) =
g ( f (x)).
We have the following formulae for domains of functions.
1. dom ( f + g) = dom f « dom g
2. dom ( f g) = dom f « dom g
3. dom (f/g) = dom f « dom g « {x : g(x) π 0}
4. dom
f = dom f « {x : f (x) ≥ 0}
Note
Note that if gof
fog
example if f : A Æ B and g : B Æ C then gof
but fog is not. Let f : R Æ R and g : R Æ R
f (x) = cos x and g(x) = x3. Then gof (x) = g( f (x)) = g(cos x)
= cos3 x and fog (x) = f (x3) = cos x3. Thus even if fog and gof
fog = gof.
DOMAINS AND RANGES OF SOME
FUNCTIONS
1. Constant functions A function that assigns the same
value to every member of its domain is called a constant
function. The domain of the constant function f (x) = c is R
and its range is {c}.
2. Polynomial functions A function of the form c xn,
where c is a constant and n is a non-negative integer, is
called a monomial in x. Examples are 2 x3, 5x4, – 6x and x8.
The function 4 x1/2 and x–3 and not monomials because the
powers of x are not non-negative integers. A function that
is expressible as the sum of finitely many monomials in x is
called a polynomial in x. Thus
f (x) =an xn + an – 1 xn – 1 + º + a1 x + a0
is a polynomial. The domain of a polynomial is R and its
range is a subset of R.
3. The domain of f (x) = loge x (= ln x) is (0, •) = {x Œ R,
x > 0} = R+, and its range is (– •, •) (i.e., the whole of R).
4. The domain of f (x) = ex is R and its range is R+.
TYPES OF FUNCTIONS
1. Rational function. This function is defined as the ratio
of two polynomials
y=
a0 x n + a1 x n - 1 + + an
b0 x m + b1 x m - 1 + + bm
For example, y = (x2 + 3)/(x3 + 4) is rational function.
2. Irrational function. If in the function y = f (x), the
operations of addition, subtraction, multiplication, division
and raising to a power with rational non-integral exponents
are performed on the right-hand side, the function y = f (x) is
said to be irrational. Examples are y = (3x2 + x )/ 2 + 4 x
and y = x .
3. Even function. A function y = f (x) is said to be an
even function if f (–x) = f (x) " x Œ dom ( f ). Examples are
y = | x |, y = cos x and y = x 2n. The graph of an even function
will be symmetric about y-axis
4. Odd function. A function y = f (x) is said to be an odd
function if f (Рx) = Рf (x) " x Πdom (f). Examples are
y = sin x and y = x2n + 1.
Clearly, y = f (x) + f (– x) is always an even function, and
y = f (x) – f (– x) is always odd. Any function y = f (x) can be
expressed uniquely as the sum of an even and an odd function as follows:
1
1
f (x) = ( f (x) + f(–x)) + ( f (x) – f (– x))
2
2
5. Periodic function. A function y = f (x) is said to be
periodic if there exists a number T > 0 such that f (x + T)
= f (x) for all x in the domain of f. The least such T is called
the period of f. For example, the period of sin x and cos x is
2p, and that of tan x is p.
If f (x) is a periodic function with period T, then the
function f (a x + b), a > 0, is periodic with period T/a. For
Sets, Relations and Functions 1.5
example, sin 2 x has a period p, cos 3x has a period 2p/3 and
tan (2 x + 4) has a period p/2. The function
Ï1 if x ŒQ
f (x) = Ì
Ó-1 if x Œ R ~ Q
is a periodic function without any period. The sum of two
periodic functions may not be periodic e.g., f (x) = {x}, the
fractional part of x and g (x) = sin x.
6. Onto function (or Surjective Function). If a function
f : A Æ B is such that each element in B is the f-image of at
least one element in A, then we say that f is a function of
A ‘onto’ B. Equivalently a function f is an onto function if
co-domain of f = Range of f.
that | f (x)| £ k for all x Œ I. Equivalently there is m and
1
M such that m £ f (x) £ M for all x Œ I. E.g. f (x) =
is
x
2
not bounded on (0, 1) and f (x) = x is bounded on [0, 1].
f (x) = sin x or cos x are bounded functions on R whereas
f (x) = tan x is unbounded on (– p /2, p /2).
GRAPHS OF SOME FUNCTIONS
1. Constant function f (x) = c represents a constant function
(Fig. 1.8).
y
y=c
(0, c )
Not onto
Onto
x
O
Fig. 1.7
For example, f : R Æ [–1, 1] defined by f (x) = sin x is
an onto function but f : R Æ R defined by f (x) = sin x
is not onto since Range of f = [– 1, 1] and co-domain of
f = R.
In order to show that a function f : A Æ B is onto we start
with any y ΠB and try to find x ΠA such that f (x) = y.
7. One-to-one function (or injective function). A function
f is said to be one-to-one if it does not take the same value
at two distinct points in its domain. For example, f (x) =
x3 is one-to-one, whereas f (x) = x2 is not, as f (1) = 1 and
f (–1) = 1. Note that a periodic function f : R Æ R cannot be
one-to-one as f (x + T) = f (x) for some T > 0.
In order to show that a function f : A Æ B is one-to-one,
we may take any x, y ΠA such that f (x) = f ( y) and try to
show that x = y.
If n(A) = m and n(B) = n(m £ n), then the number of injections (or one-one functions) from A to B is
n
n!
.
Pm =
(n - m)!
8. Bijective function (One-to-One and Onto). If a function
f is both one-to-one and onto, then f is said to be a bijective function. For example an identity function iA : A Æ A
defined by iA (x) = x is trivially a bijective function. The
function f : [–p/2, p/2] Æ [–1, 1] defined by f (x) = sin x
is a bijective function. The function f : (– p/2, p/2) Æ R
defined by f (x) = tan x is also a bijective function.
If A and B are finite sets and f: A Æ B is a bijection then
n(A) = n(B). If n(A) = n, then the number of bijections from
A to B is the total number of arrangements of n items taken
all at a time which is n!
9. Bounded and unbounded functions A function f defined
on an interval I is said bounded on I if there is k > 0 such
Fig. 1.8
2. Proportional values. If variables y and x are direct proportional, then the functional dependence between them is
represented by the equation:
y = kx,
where k is a constant a factor of proportionality.
A graph of a direct proportionality is a straight line,
going through an origin of coordinates and forming with
an x-axis an angle a, a tangent of which is equal to k:
tan a = k. Therefore, a factor of proportionality is called
also a slope. These are shown in three graphs with k = 1/3,
k = 1 and k = –3 on Fig.1.9
k=–3
Y
k=1
k = 1/3
a
O
X
Fig. 1.9
3. Linear function. If variables y and x are related by the
1-st degree equation:
Ax + By = C,
(at least one of numbers A or B is non-zero), then a
graph of the functional dependence is a straight line. If
C = 0, then it goes through an origin of coordinates, otherwise – not. Graphs of linear functions for different combinations of A, B, C are represented on Fig.1.10
1.6
Complete Mathematics—JEE Main
parabola – a curve going through an origin of coordinates.
Every parabola has an axis of symmetry OY, which is called
an axis of parabola. The point O of intersection of a parabola
with its axis is a vertex of parabola.
Y
y = 1/2 x2
3
2
1
Fig. 1.10
4. Inverse proportionality. If variables y and x are inverse
proportional, then the functional dependence between them
is represented by the equation:
y = k/x,
where k is a constant.
A graph of an inverse proportionality is a curve, having
two branches (Fig. 1.11). This curve is called a hyperbola.
These curves are received at crossing a circular cone by a
plane. As shown on Fig. 1.11, a product of coordinates of a
hyperbola points is a constant value, equal in this case to 1.
In general case this value is k, as it follows from a hyperbola
equation:
xy = k
Y
–3 –2 –1 O
–1
1
2
3
X
Fig. 1.12
A graph of the function y = ax2 + bx + c is also a quadratic
parabola of the same shape, that y = ax2, but its vertex is not
an origin of coordinates, this is a point whose coordinates:
Ê b
b2 ˆ
c
,
ÁË 2a
4a ˜¯
The form and location of a quadratic parabola in a coordinate system depends completely on two parameters: the
coefficient a of x2 and discriminant D : D = b2 – 4ac. These
properties follow from analysis of the quadratic equation
roots. All possible different cases for a quadratic parabola
are shown on Fig. 1.13
2
1
–2
–1
X
O
–1
–2
Fig. 1.11
The main characteristics and properties of hyperbola:
∑ the function domain: x π 0, and codomain: y π 0;
∑ the function is monotone (decreasing) at x < 0 and at
x > 0, but it is not monotone on the whole, because of
a point of discontinuity x = 0
∑ the function is unbounded, discontinuous at a point
x = 0, odd, non-periodic;
∑ there are no zeros of the function.
5. Quadratic function. This is the function: y = ax2 + bx + c,
where a, b, c – constants, a π 0. In the simplest case we have
b = c = 0 and y = ax2. A graph of this function is a quadratic
Fig. 1.13
The main characteristics and properties of a quadratic
parabola:
∑ the function domain: – • < x < + • (i.e. x is any real
number)
∑ the function is not monotone on the whole, but to the
right or to the left of the vertex it behaves as a monotone function;
∑ the function is unbounded, continuous in everywhere,
even at b = c = 0, and non-periodic;
∑ the function has no zeros at D < 0.
Sets, Relations and Functions 1.7
6. Power function. This is the function: y = axn where a,
n are constants. At n = 1 we obtain the function, called a direct
proportionality: y = ax; at n = 2 – a quadratic parabola; at
n = –1 – an inverse proportionality or hyperbola. So, these
functions are particular cases of a power function. We know,
that a zero power of every non-zero number is 1, thus at
n = 0 the power function becomes a constant: y = a, i.e.
its graph is a straight line, parallel to an x-axis, except an
origin of coordinates. All these cases (at a = 1) are shown on
Fig. 1.14 (n ≥ 0) and Fig. 1.15 (n < 0).
Negative values of x are not considered here, because
then some of functions:
that this is the two-valued function (the sign ± before the
square root symbol says about this). Such functions are not
studied in an elementary mathematics, therefore we consider usually as a function one of its branches: either an
upper or a lower branch.
Y
y = x2
O
X
y = x1/2, y = x1/4
y = x3
lose a meaning.
Y
2
n=4 n=2
n=1
Fig. 1.16
n
y = x ,n > 0
Y
n = 1/2
n = 1/4
n=0
1
y=± x
O
O
1
2
X
Fig. 1.17
Fig. 1.14
Y
2 – 1/2 – 1 – 2
y = xn,n <0
1
n = – 1/2
n=–1
n=–2
O
1
X
2
X
Fig. 1.15
If n – integer, power functions have a meaning also at
x < 0, but their graphs have different forms depending on
that is n an even or an odd number. In Fig. 1.16 two such
power functions are shown: for n = 2 and n = 3.
At n = 2 the function is even and its graph is symmetric
relatively an axis Y; at n = 3 the function is odd and its graph
is symmetric relatively an origin of coordinates. The function y = x3 is called a cubic parabola.
On Fig. 1.17 the function y = ± x is represented. This
function is inverse to the quadratic parabola y = x2, its graph
is received by rotating the quadratic parabola graph around
a bisector of the 1-st coordinate angle. We see by the graph,
7. Exponential function. The function y = ax, where a is a
positive constant number, is called an exponential function.
The argument x adopts any real values; as the function values
only positive numbers are considered, because otherwise we
will have a multi-valued function. So, the function y = 81x
has at x = 1/4 four different values: y = 3, y = –3, y = 3i and
y = – 3i. But we consider as the function value only y = 3.
Graphs of an exponential function for a = 2 and a = 1/2 are
shown on Fig. 1.18. All they are going through the point (0, 1).
At a = 1 we have as a graph a straight line, parallel to x-axis,
i.e. the function becomes a constant value, equal to 1. At a > 1
an exponential function increases, and at 0 < a < 1 – decreases.
Y
a = 1/2
a=2
5
3
1
–3 –2 –1
1
Fig. 1.18
2
3
X
Complete Mathematics—JEE Main
y = ex
y=
y = e|x|
e–x
Fig. 1.19
Fig. 1.20
y
y = |x|
y=
-
x
x
0
Fig. 1.22
Greatest Integer Function and Fractional part function
f(x) = [x] will denote the integral part of x or greatest integer
less than or equal to x. E.g. [5] = 5, [5.2], = 5, [–5.1] = – 6,
[p] = 3, [e] = 2. The domain of this function is R and the range is I.
For positive x, this function might represent, for example, the
legal age of a person is a function of his chronological age x.
The fractional part of x is the fractional part {x} of x so
{x} = x – [x] e.g. {5.04} = 0.04, {–4} = 0, {–1.7} = 0.3, {p} = p
– 3. The domain of this function is R whereas the range is [0, 1).
y
8. Logarithmic function. The function y = logax, where a is
a positive constant number, not equal to 1 is called a logarithmic function. This is an inverse function relatively to an
exponential function; its graph (Fig.1.19) can be obtained
by rotating a graph of an exponential function around of a
bisector of the 1-st coordinate angle.
3
2
2
1
1
0
x¢
Y
x
The main characteristics and properties of an exponential
function:
∑ the function domain: –– • < x < + • (i.e. x is any real
number) and its codomain: y > 0;
∑ this is a monotone function: it increases at a > 1 and
decreases at 0 < a < 1;
∑ the function is unbounded, continuous in everywhere,
non-periodic;
∑ the function has no zeros.
y=
1.8
1
a>1
1
2
x
3
graph of [x]
2
3
O
X
y¢
1
Fig. 1.23
y
0< a <1
Fig. 1.21
The main characteristics and properties of a logarithmic
function:
∑ the function domain: x > 0 and its codomain: – • < y
< + • (i.e. y is any real number);
∑ this is a monotone function: it increases for a > 1 and
decreases for 0 < a < 1;
∑ the function is unbounded, continuous in everywhere,
non-periodic;
∑ the function has one zero: x = 1.
9. Absolute-value function is given by
Ï x , x≥0
y = | x | = max {x, – x} = Ì
Ó- x , x < 0
It is depicted in Fig. 1.22.
1
x¢
2
3
x
graph of {x}
y¢
Fig. 1.24
Some Properties
1.
2.
3.
4.
x – 1 < [x] £ x
[{x}] = 0 ={[x]}
If n £ x < n + 1, n Œ I, then [x] = n and conversely.
If x ΠI, then [x] = x otherwise [x] < x.
, if x ΠI
Ï-[ x ]
5. [–x] = Ì
Ó-[ x ] - 1 , if x œ I
Ï0, if x Œ I
in other word, [ x ] + [ - x ] = Ì
Ó-1, if x œ I
Sets, Relations and Functions 1.9
6. [x + y] = [x] + [y], if x or y ΠI.
Figure 1.26 shows the graph. For positive value of x, this
function might represent for example, the cost of parking x
hours in a parking lot which charges Rs 10 for each hour or
part of an hour in this case f (x) = 10 [x]¢.
if {x} + {y} < 1
Ï[ x ] + [ y],
7. [x + y] = Ì
Ó[ x ] + [ y] + 1, if {x} + {y} ≥ 1
Moreover, [x + y] ≥ [x] + [y].
Ï x,
8. {x} = Ì
Ó0,
INVERSE OF A FUNCTION
if 0 < x < 1
x ŒI
The function f (x) = x1/3 and g(x) = x3 have the following
property:
f (g(x)) = f (x3) = (x3)1/3 = x
g( f (x)) = g(x1/3) = (x1/3)3 = x
Similarly, the functions f (x) = loge x and g(x) = ex cancel
the effect of each other. If two functions f and g satisfy
f (g(x)) = x for every x in the domain of g, and g( f (x)) = x
for every x in the domain of f, we say that f is the inverse of
g and g is the inverse of f, and we write f = g–1, or g = f –1.
(The symbol f –1 does not mean 1/f.) To find the inverse
of f, write down the equation y = f (x) and then solve x as a
function of y. The resulting equation is x = f –1 ( y).
If f is one-to-one, then f has an inverse defined on its
range and, conversely, if f has an inverse, then f is one-toone. f –1 is defined on the range of f. If f is one-to-one from A
to B, and g is one-to-one from B to C, then f o g is one-to-one
from A to C, and ( f o g)–1 = g–1 o f –1.
If f : X Æ Y is bijective then f –1 : Y Æ X is bijective and
Ï0, x Œ I
9. {x} + {–x} = Ì
Ó1, if x œ I
n - 1˘
1˘
È
È
10. [ x ] + Í x + ˙ + ... + Í x +
= [n x ] .
n˚
n ˙˚
Î
Î
11. Signum function is given by
if x > 0
Ï1
Ô
sgn (x) = Ì0 if x = 0
Ô-1 if x < 0
Ó
and it is depicted in Fig. 1.25
y
y = sgn(x )
0
x
1
( f -1 )
12. The least integer function: The function whose
value at any number x is the smallest integer greater
than or equal to x is called the least integer function, e.g. if
f (x) = [x]¢ = smallest integer, greater than or equal
to x.
then [1.5]¢ = 2 [1.99]¢ = 1 [1.4]¢ = 2.
y
1
9
{ }
ax + b
d
. The domain of f is R ~ .
cx + d
c
ax + b
b - dy
so cxy + dy = ax + b fi x =
.
Let y =
cx + d
cy - a
b - dx
Hence f –1(x) =
.
cx - a
Let f(x) =
DIRECT AND INVERSE IMAGES
y = [(x]
2
= f. Moreover, f –1 ({ y }) = {f –1( y )} for all y Œ Y.
Illustration
Fig. 1.25
3
-1
1
2
3
4
x
Let f be a function with domain X and co-domain Y. If
A Õ X, then the direct image of A under f is the subset of Y
(denoted by f (A)) is defined to be { f (x) : x ΠA}.
For example, let f : R Æ R be a function defined by
f (x) = x2. If A = {–3, –1, 0, 1, 3}, then f (A) = {0, 1, 9}.
Note that if f (X) = Y, then the function f : X Æ Y is onto.
The following relations can be verified easily.
(i) A Ã B fi f (A) Ã f (B)
(ii) f (A » B) = f (A) » f (B)
(iii) f (A « B) Ã f (A) « f (B). The inclusion may be strict.
Fig. 1.26
Let f be a function with domain X and range Y and let B
be a subset of Y. The inverse image of B under f is a subset
of X (denoted by f –1 (B)) is defined to be {x : f (x) Œ B}.
1.10
Complete Mathematics—JEE Main
Illustration
Note that if f is one-to-one, then f –1({ x }) is a singleton set.
The following relations can be verified easily for A, B Ã Y.
10
If f : R Æ R is a function defined by f (x) = | x | and A = (– •,
0), then f –1(A) = {x: f (x) Œ A} = {x: | x | Œ (– •, 0)} = f (the
empty set). If B = (–3, 3), then f –1 (B) = (–3, 3).
(i) A Ã B = f –1 (A) Ã f –1 (B)
(ii) f –1 (A » B) = f –1 (A) » f –1 (B)
(iii) f –1 (A « B) = f –1 (A) « f –1 (B).
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Example 1: Let I be set of integers, N = the set of nonnegative integers, Np = the set of non-positive integers. Then
the sets A and B satisfying A « B = f are
(a) A = I ~ Np, B = N ~ Np
(b) A = I ~ N, B = I ~ Np
(c) A = N D Np, B = I ~ Np
(d) A = N D Np, B = (I ~ N) » {0}
Ans. (b)
Solution: I ~ N = {… –3, –2, –1}, I ~ Np = {1, 2, 3…},
N ~ Np = {1, 2, …}, N D Np = (N ~ Np) » (Np ~ N)
= {0, 1, 2, …}
(I ~ N) » {0} = {…, –3, –2, –1, 0}
The disjoint sets are I ~ N and I ~ Np.
Example 2: Which of the following equality is not true.
(a) A « (B ~ C) = A « B ~ (A « C)
(b) A ~ (A « B) = A ~ B
(c) A ~ (B ~ C) = (A ~ B) » (A « C)
(d) A ~ (B D C ) = (A ~ B) D (A ~ C)
Ans. (d)
Solution: For equality (a),
A « (B ~ C) = A « (B « C¢)
= f » (A « B « C¢)
= (A « B « A¢) » (A « B « C¢)
= A « B « (A¢ » C¢)
= A « B ~ (A « C)
For equality (b),
A ~ (A « B) = A « (A¢ » B¢)
= (A « A¢) » (A « B¢)
= f » (A « B¢) = A « B¢ = A ~ B
For equality (c)
A ~ (B ~ C) = A ~ (B « C¢) = A « (B¢ » C)
= (A « B¢) » (A « C)
= (A ~ B) » (A « C)
For (d) Let A = {1, 2, 3, 4, 5}, B = {3, 4, 5}, C = {1, 2, 3}
So,
B D C = {4, 5} » {1, 2} = {1, 2, 4, 5}
Thus A ~ (B D C) = {3}
A ~ B = {1, 2}; A ~ C = {4, 5}
Therefore (A ~ B) D (A ~ C) = ({1, 2} ~ {4, 5})
» ({4, 5} ~ {1, 2})
= {1, 2} » {4, 5} = {1, 2, 4, 5).
Hence A ~ (B D C) π (A ~ B) D (A ~ C)
Example 3: A boating club consists of 82 members,
each member is either a sailboat owner or a powerboat
owner. If 53 members owned sailboats and 38 members
owned powerboats, the number of members owned both
sailboat and powerboat is
(a) 6
(b) 7
(c) 9
(d) 4
Ans. (c)
Solution: Let S = the set of all members owning sailboats and P = the set of all members owning powerboats
n(S » P) = n(S) + n(P) – n(S « P)
82 = 53 + 38 – n(S « P)
fi n(S « P) = 91 – 82 = 9.
Example 4: If, B Ã A¢, then which of the following sets
π
is equal to A¢.
(a) (A « B) » B
(b) (A « B) » A¢
(c) (A » B) « A¢
(d) (A » B) « B
Ans. (b)
Solution: For (a), (A « B) » B = (A » B) « (B » B)
= (A » B) « B = B.
For (b), (A « B) » A¢ = (A » A¢) « (B » A¢)
= X « A¢ = A¢
For (c), (A » B) « A¢ = (A « A¢) » (B « A¢)
= f » (B « A¢)
= B « A¢ = B.
For (d) (A » B) « B = B.
È x ˘ È x ˘ 5x
Example 5: If Í ˙ + Í ˙ = , then x is any term of
Î2˚ Î3˚ 6
the following
(a) 3, 6, 9, 12,…
(c) 6, 12, 18, 24,…
Ans. (c)
(b) 9, 18, 27, 36, …
6 12 18
(d) , , ,...
5 5 5
Sets, Relations and Functions 1.11
Èx˘ Èx˘
Èx˘ Èx˘
Solution: Since Í ˙ , Í ˙ Œ I , so Í ˙ + Í ˙ Œ I
Î2˚ Î3˚
Î2˚ Î3˚
5x
6
ΠI fi x = n, n ΠI .
Thus
6
5
È x ˘ È x ˘ 5x
Substituting this value in Í ˙ + Í ˙ = , we have
Î2˚ Î3˚ 6
È3 ˘ È2 ˘
ÍÎ 5 n ˙˚ + ÍÎ 5 n ˙˚ = n
{ } { }
{ }{ }
{ } { }
3
3
2
2
n- n + n- n =n
5
5
5
5
3
2
n + n =0
5
5
fi
fi
3
2
n =0= n
5
5
fi
(Since 0 £ {x} < 1)
2n ◊ 3n
5
5m1 ¥ 5m2
x ◊ 5m1
=
= 5m1 m2 fi
= 5m1 m2
5
3
fi
x = 3m2
x ◊ 5m2
Similarly
= 5m1 m2 fi x = 2 m1
2
Hence x is multiple of 2 and 3 so of 6 and x ΠI
Thus 3n = 5m1, 2n = 5m2, Therefore xn =
Example 6: The relation R defined by ‘>’ on the set N is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence relation
Ans. (c)
</ 3
Solution: 2 >/
so (3, 2) Œ R but (2, 3) œ R. Thus R is not symmetric. If
(a, b) ΠR and (b, c) ΠR then a > b, b > c fi a > c so (a, c)
ΠR. R is not an equivalence relation.
Example 7: The relation a R b defined by a is factor of
b on N is not
(a) reflexive
(b) transitive
(c) anti symmetric
(d) symmetric
Ans. (d)
Solution: For a ΠN, a is a factor of a so R is reflexive.
If a is factor b and b is factor of c then a is factor of c so R is
transitive. If a is factor of b and b is factor of a then a = b so
R is anti symmetric, 2 is factor of 4 but 4 is not a factor of 2.
Example 8: The domain of the function f(x) = log2 sin x is
(a) R
(b) R ~ {n p : n ΠI}
(c) R ~ {np : n ΠN}
(d) » (2np , (2n + 1)p )
nŒI
Ans. (d)
Solution: f(x) = log2 sin x is defined for all x for
which sin x > 0. But sin x > 0 if x Œ (0, p) » (2p, 3p) » …
= » (2np , (2n + 1)p ) .
nŒI
Example 9: The domain of y = cos-1
È 3 5˘
(a) Í - , ˙
Î 2 2˚
(c) [0, 2]
Ans. (a)
1 - 2x
is
4
(b) [–1, 1]
È 1 3˘
(d) Í - , ˙
Î 2 2˚
Solution: The given function is defined if
1 - 2x
-1 £
£ 1 i.e. if - 4 £ 1 - 2 x £ 4 ,
4
5
-3
fi – 5 £ – 2x £ 3 fi
£x≥
2
2
Example 10: Which of the following functions is bounded
(a) y = 1 – log10x
(b) y = e–2x
–1
(c) y = sin (2x + 1)
(d) y = tan (4x + 1)
Ans. (c)
Solution: The range log10x is (–•, • ) so y = 1 – log10x
is unbounded. y = e–2x is unbounded from below as x Æ – •,
È -p p ˘
yÆ •. The range of sin–1x is Í , ˙ so y = sin–1(2x + 1) is
Î 2 2˚
a bounded function. The range of tan x is R so y = tan (4x + 1)
is unbounded
Example 11: A function out of the following whose
period is not p is
(a) sin2x
(b) cos2x
(c) tan (2x + 3)
(d) y = |sin x|
Ans. (c)
1
Solution: y = sin2x = [1 - cos 2 x ] . Since period of
2
cos x is 2p so period of cos 2x is p.
1
y = 1 + cos2x = 1 + (1 + cos 2 x ) . The period of this is
2
again p. The period of tan x is p so period of tan (2p + 3) is
p/2. If f(x) = |sin x| then f(x + p) = |sin (x + p)| = |–sin x| =
Ásin x Á = f(x). Thus the period of f is p.
Example 12: Which of the following functions is an odd
function
(a) y = x4 – 2 x2
(b) y = x –x2
3
5
(c) y = cos x
(d) y = x - x + x
6 40
Ans. (d)
Solution: If f(x) = x4 – 2x2 then f(–x) = (–x)4 –2(–x)2 =
x4 – 2x2 = f(x). Hence f is an even function. If u(x) = x – x2
then u(–x) = x – x2 so u is neither even nor an odd function.
p(x) = cos x, p(–x) = cos (–x) = cos x = p(x) so p is even
x3 x 5
x3 x 5
+
then s(–x) = - x =
function. If s(x) = x 6 40
6 40
3
5
Ê
x
x ˆ
-Á x + ˜ = – s(x), so s is an odd function.
6 40 ¯
Ë
1.12
Complete Mathematics—JEE Main
LEVEL 1
Straight Objective Type Questions
Example 13: Let X = {x : x = n3 + 2n + 1, n ΠN} and
Y = {x : x = 3n2 + 7, n ΠN} then
(a) X « Y is a subset of {x : x = 3n + 5, n Œ N}
(b) X « Y Õ {x : x = n2 + n + 1, n Œ N}
(c) 34 Œ X « Y
(d) none of these
Ans. (c)
n
r,
.
Solution: If n3 + 2n + 1 = 3n2 + 7
fi
n3 – 3n2 + 2n – 6 = 0
fi
(n – 3) (n2 + 2) = 0
fi
n = 3 as n ΠN
So,
x = 3 ¥ 32 + 7 = 34 Œ X « Y.
In (a) and (b) x π 34, for any n Œ N.
Example 14: A, B, C are the sets of letters needed to spell
the words STUDENT, PROGRESS and CONGRUENT,
respectively. Then n (A » (B « C )) is equal to
(a) 8
(b) 9
(c) 10
(d) 11
Ans. (b)
Solution:
A=
B=
C=
So
A » (B « C)
=
=
and n [(A » (B « C)]
{D, E, N, S, T, U ]
{E, G, O, P, R, S }
{C, E, G, N, O, R, T, U }
{D, E, N, S, T, U } » {E, G, O, R}
{D, E, G, O, N, R, S, T, U }
= 9.
Example 15: Let A = {x : x is a prime factor of 240}
B = {x : x is the sum of any two prime factors of 240}.
Then
(a) 5 œ A « B
(b) 7 Œ A « B
(c) 8 Œ A « B
(d) 8 Œ A » B
Ans. (d)
Solution: 240 = 2 ¥ 3 ¥ 5 ¥ 8
So
A = {2, 3, 5}, B = {5, 7, 8}.
Clearly 8 Œ A » B.
Example 16: A, B, C are three sets such that n(A) = 25,
n(B) = 20, n(c) = 27, n(A « B) = 5, n(B « C) = 7 and A « C =
f then n(A » B » C) is equal to
(a) 60
(b) 65
(c) 67
(d) 72.
Ans. (a)
Solution: A « C = f fi A « B « C = f.
\ n(A » B » C) = n(A) + n(B) + n(C)– n(A « B) – n(B « C)
– n(A « C) + n(A « B « C)
= 25 + 20 + 27 – 5 – 7 – 0 + 0 = 60.
Example 17: Let X = {(x,y,z) | x,y,z ΠN. x + y + z =
10, x < y < z} and Y = {(x,y,z) | x,y,z ΠN, y = |x Рz|} then
X « Y is equal to
(a) {(2,3,5)}
(b) {1,4,5}
(c) {5,1,4}
(d) {(2,3,5), (1,4,5)}
Ans. (d)
Solution: X = {(1,2,7), (1,3,6), (1,4,5), (2,3,5)}.
Elements of X which belong to Y are (1,4,5) and (2,3,5) both
so they belong to X « Y.
Example 18: If A,B,C are three non-empty sets such
that A « B = f, B « C = f, then
(a) A = C
(b) A Ã C
(c) C Ã A
(d) none of these
Ans. (d)
Solution: Let A = {1,2,3,4,5}, B = {6,7,8,9} and
C = {11,12,13} which satisfy the given conditions but none
of (a), (b) or (c).
Example 19: Two finite sets have m and n elements
respectively. The total number of subsets of first set is 56
more than the total number of subsets of the second set. The
values of m and n respectively are
(a) 7, 6
(b) 6, 3
(c) 5, 1
(d) 8, 7
Ans. (b)
Solution: According to the given condition, we have
2m = 2n + 56
fi 2m –3– 2n –3 = 7 fi 2n –3 (2m – n – 1) = 7. Since 7 is a prime
number so we must have n – 3 = 0 (clearly m π n). Thus
n = 3. Therefore, 2m = 23 + 56 = 64 = 26 fi m = 6.
Example 20: Among employee of a company taking
vacations last years, 90% took vacations in the summer, 65%
in the winter, 10% in the spring, 7% in the autumn, 55% in
winter and summer, 8% in the spring and summer, 6% in
the autumn and summer, 4% in the winter and spring, 4% in
winter and autumn, 3% in the spring and autumn, 3% in the
summer, winter and spring, 3% in the summer, winter and
autumn, 2% in the summer, autumn and spring, and 2% in
the winter, spring and autumn. Percentage of employee that
took vacations during every season:
Sets, Relations and Functions 1.13
(a) 4
(c) 2
Ans. (c)
(b) 3
(d) 8
Solution: Suppose that number of employee taking vacations is 100.
Su – set of employee taking leave in Summer
W – set of employee taking leave in Winter
Sp – set of employee taking leave in Spring
A – set of employee taking leave in Autumn
n(Su) = 90, n(W) = 65, n(Sp) = 10, n(A) = 7
n(W « Su) = 55, n(Sp « Su) = 8, n(A « Su) = 6
n(W « Sp) = 4, n(W « Au) = 4, n(Sp « A) = 3
n(Su « A) = 3, n(Su « W « A) = 3
n (Su « W « Sp) = 3, n(Su « A « Sp) = 2
n(W « Sp « A) = 2
n(Su « Sp « W « A)
= n(Su) + n(Sp) + n(W) + n(A) – n(Su « Sp)
– n(Sp « W) – n(W « A) – n(Su « A) – n(Su « W)
– n(Sp « A) + n(Su « Sp « W) + n(Su « W « A)
+ n(W « A « Su) + n(Su « Sp « A)
–n(Sp » Su » A » W)
= 90 + 65 + 10 + 7 – 55 – 8 – 6 – 4 – 4 – 3
+ 3 + 3 + 2 + 2 – 100 = 2
Example 21: If A = {1, 2, 3, 4}, B = {3, 4, 5} then the
number of elements in (A » B) ¥ (A « B) ¥ (A D B) is
(a) 5
(b) 30
(c) 10
(d) 4
Ans. (b)
Solution: A » B = {1, 2, 3, 4, 5}. n(A » B) = 5
A « B = {3, 4}, n(A « B) = 2
A D B = (A ~ B) » (B ~ A) = {1, 2} » {5}
= {1, 2, 5}
n(A D B) = 3. Hence n((A » B) ¥ (A « B) ¥ (A D B)}
= 5 ¥ 2 ¥ 3 = 30.
Example 22: Let I be the set of integers. For a, b ΠI,
a R b if and only if |a – b| < 1, then
(a) R is not reflexive
(b) R is not symmetric
(c) R = {(a,a); a ΠI}
(d) R is not an equivalence relation.
Ans. (c)
Solution: For any integers a, b, |a – b| < 1 if and only
if |a Рb| = 0 so a = b. Hence R = {(a,a); a ΠI}.Thus R is
reflexive, symmetric and transitive.
Example 23: Let W denote the words in the English
Dictionary. Define the relation R by R = {(x, y) Œ W ¥ W :
the words x and y have at least one letter common}, then R is
(a)
(b)
(c)
(d)
Ans. (c)
reflexive, not symmetric and transitive
not reflexive, symmetric and transitive
reflexive, symmetric and not transitive
reflexive, symmetric and transitive
Solution: (x, x) ΠR V x ΠW as all letters in both
are common. If (x, y) ΠR then x and y have a letter in common fi (y, x) ΠR.
Next, let x = fix, y = six and z = son then(x, y) ΠR,
(y, z) Œ R but (x, z) œ R
So R is reflexive, symmetric but not transitive
Example 24: If the relation R : A Æ B, where A = {1, 2, 3, 4}
and B = {1, 3, 5} is defined by R = {(x, y); x < y, x Œ A, y Œ
B} then RoR –1 is
(a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
(b) {(3, 1), (5, 1), (5, 2), (5, 3), (5, 4)}
(c) {(3, 3), (3, 5), (5, 3), (5, 5)
(d) none of these
Ans. (c)
Solution: R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
and
R –1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}.
Thus
RoR –1 = {(3, 3), (3, 5), (5, 3), (5, 5)}.
Example 25: Let A = {x Œ R : [x + 3] + [x + 4] £ 3} and
x -3
•
ÏÔ
¸Ô
3 ˆ
xÊ
B = Ìx ŒR : 3 Á Â r ˜
< 3-3 x ˝ then
Ë r =1 10 ¯
ÓÔ
˛Ô
(a) A = B
(b) A Ã B
(c) B Ã A
(d) A « B = f
π
π
Ans. (a)
Solution: Let x Œ A, [x + 3] + [x + 4] £ 3
fi
[x] + 3 + [x] + 4 £ 3
fi
2[x] £ – 4 fi [x] £ –2
fi x Œ (– •,–1)
A = (– •, –1)
x
If x ΠB then 3 3
x–3 Ê
•
1 ˆ
ÁË Â 10r ˜¯
r =1
fi
Ê 1 10 ˆ
32 x -3 Á
Ë 1 - 1 10 ˜¯
fi
32 x -3 3-2
( )
x -3
x -3
< 3-3 x
x -3
< 3-3 x
< 3-3 x
33 < 3–3x fi 3 < –3x
so x Œ (– •, –1)
Hence B = (– •, –1). Thus A = B.
fi
Example 26: The range of the function f(x) = 7 – x Px – 3 is
(a) {1, 2, 3, 4}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3}
(d) {1, 2, 3, 4, 5}
Ans. (c)
1.14
Complete Mathematics—JEE Main
Solution: 7 – x ≥ 1, x – 3 ≥ 0
and
7–x≥x–3
fi
x £ 6, x ≥ 3, x £ 5.
Thus
3£x£5
\
Range = {4P0, 3P1, 2P2}
= {1, 3, 2}
Example 27: The domain of the function
f(x) =
sin -1 ( x - 3)
9 - x2
(a) [1, 2]
(c) [1, 3)
Ans. (b)
is
Solution: Since x – x + 2 = 2 which is an irrational
number so x R x for all x ΠR. Hence R is reflexive. R is
not symmetric as ( 2 , 1) Œ R but (1, 2 ) œ R. Again R
is not transitive since ( 2 , 1) ŒR and (1, 2 2 ) ŒR but
( 2 , 2 2 ) œR.
Example 32: If n Áq and A = {z ŒC : zn = 1},
B = {z : zq = 1} then
(a) A = B
(b) A « B = {1}
(c) B Ã A
(d) A Ã B
π
(b) [2, 3)
(d) [1, 2]
Solution: – 1 £ x – 3 £ 1 and 9 – x2 > 0
fi
2 £ x £ 4 and – 3 < x < 3.
So domain of f is [2, 3).
Example 28: The solution of 8x ∫ 6 (mod 14) is
(a) [8], [6]
(b) [6], [14]
(c) [6], [13]
(d) [8], [14], [16]
where [a] = {a + 14 k : k ΠI}
Ans. (c)
Solution: We need to solve 14 Á(8x – 6) i.e., 8x – 6 =
14k, for k ΠI. The values 6 and 13 satisfy this equation,
while 8, 14 and 16 do not.
Example 29: Let R = {(x, y): x, y ΠA, x + y = 5}, where
A = {1, 2, 3, 4, 5} then
(a) R is not reflexive, symmetric and not transitive
(b) R is an equivalence relation
(c) R is reflexive, symmetric but not transitive
(d) R is not reflexive, not symmetric but transitive
Ans. (a)
Solution: R = {(1, 4), (4, 1), (2, 3), (3, 2)}, so R is not
reflexive as (1, 1) œ R. R is symmetric by definition and R is
not transitive as (1, 4) ŒR, (4, 1) Œ R but (1, 1) œR.
Example 30: Let R be a relation on a set A such that
R = R –1 then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) an equivalence relation
Ans. (b)
Solution: If (a, b) Œ R then (b, a) Œ R–1= R so R is
symmetric. The relation in Example 29 satisfy R = R–1 but
is neither reflexive nor transitive.
Example 31: For x, y ΠR, define a relation R by x R y
if and only if x – y + 2 is an irrational number. Then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) none of these
Ans. (a)
Ans. (d)
Solution: q = p n for some p ΠN
z q – 1 = (z n)p –1 = (z n – 1) (z(p – 1)n + ... + zn + 1)
Every root of z n –1 is a root of zq – 1 and every root of
(p–1)n ...
z
+ + zn + 1 = 0 is also a root of zq – 1. Hence A Ã B
π
and A « B = A.
Example 33: If A = {z : (1 + 2i) z + (1 – 2i) z + 2 = 0}
and B = {z : (3 + 2i) z + (3 – 2i) z + 3 = 0}, then
(a) A « B is a singleton set
(b) A Ã B
(c) B Ã A
(d) A « B = f.
Ans. (a)
Solution: Equations in sets A and B represent straight
a
a
line with a1 = 1 +2i and a 2 = 3 + 2i. Since 1 π 2 so the
a1 a 2
lines are intersecting, hence A « B is a singleton set.
Example 34: Let x, y ŒI and suppose that a relation R
on I is defined by x R y if and only if x £ y then
(a) R is reflexive but not symmetric
(b) R is an equivalence relation
(c) R is neither reflexive nor symmetric
(d) R is symmetric but not transitive
Ans. (a)
Solution: Since x £ x for all x Œ I so R is reflexive but is
not symmetric as (1, 2) Œ R and (2, 1) œ R. Also R is transitive as x £ y, y £ z fi x £ z.
Example 35: If f : R Æ R is defined by f(x) = x2 + 1,
then value of f –1 (17) and f –1 (–3) are, respectively,
(a) f, {4, – 4}
(b) f, {3, –3}
(c) {3, –3}, f
(d) {4, – 4}, f
Ans. (d)
f
–1
Solution: For any A Õ R, we have
(A) = {x ΠR: f (x) ΠA}. Thus,
f –1 (17) = {x : f (x) Œ {17}} = {x : f (x) = 17}
= {x : x2 + 1 = 17} = {4, – 4},
and similarly, f –1 (– 3)= {x Œ R : x2 + 1 = – 3} = f.
Sets, Relations and Functions 1.15
Example 36: The functions f and g are given by f(x) = {x},
1
the fractional part of x and g(x) =
sin [x]p, where [x]
2
denotes the integral part of x. Then range of gof is
(a) [–1, 1]
(b) {0}
(c) {–1, 1}
(d) [0, 1]
Ans. (b)
Solution: (gof) (x) = g ( f(x)) = 1/2 sin [{x}]p = 0, for
all x ΠR. Hence the range of gof is {0}.
Example 37: The period of the function
f (x) = cos2 3x + tan 4x is
(a) p/3
(b) p/4
(c) p/6
(d) p
Ans. (d)
Solution: f (x) = (1/2) (1 + cos 6x) + tan 4x. The period
of cos 6x is 2p/6 = p/3 and the period of tan 4x is p/4. Hence
the period of f is l.c.m. of p/3 and p/4 = p.
Example 38: The domain of the function
Ê
f (x) = sin–1 Á log3
Ë
(a) [–1, 9]
(c) [–9, 1]
Ans. (b)
xˆ
˜ is
3¯
(b) [1, 9]
(d) [3, 9]
Solution: The function f is defined only if
- 1 £ log3 (x/3) £ 1. This inequality is possible only if
1/3 £ x/3 £ 3 i.e., 1 £ x £ 9.
Example 39: The domain of the function
f(x) =
- log0.3 ( x - 1)
- x2 + 2 x + 8
(a) (1, 4)
(c) (2, 4)
Ans. (c)
is
(b) (–2, 4)
(d) none of these
Solution: Since for, 0 < a < 1, loga x < 0 for x > 1
so log0.3 (x - 1) < 0 for x > 2. Also - x2 + 2x + 8 > 0 if
and only if x Π(- 2, 4). Hence the domain of the given
function is (2, 4).
Example 40: The function f: [–1/2, 1/2] Æ [–p/2, p/2]
defined by f(x) = sin–1 (3x – 4x3) is
(a) both one-one and onto
(b) neither one-one nor onto
(c) onto but not one-one
(d) one-one but not onto
Ans. (a)
Solution: Since sin–1 (3x – 4x3) = 3sin–1 x Œ [–p/2, p/2]
i.e., sin–1 x Œ [–p/6, p/6] or x Œ [–1/2, 1/2] so f is onto.
3
> 0 for –1/2 < x < 1/2. Therefore, f
Also f ¢( x ) =
1 - x2
increases on [–1/2, 1/2] and hence f is one-one.
1
Example 41: Given f(x) =
1
g(x) =
x-|x|
(a)
(b)
(c)
(d)
Ans. (a)
|x|- x
and
then
dom f π f and dom g = f
dom f = f and dom g π f
f and g have the same domain
dom f = f and dom g = f
Solution: dom f = {x : | x | > x} and dom g = {x : x > | x |}
= f. Thus dom f = R– (the set of all negative real numbers)
and dom g = f.
Example 42: Which of the following functions is not
onto
(a) f: R Æ R, f(x) = 3x + 4
(b) f: R Æ R+, f (x) = x2 + 2
(c) f: R+ Æ R+, f(x) =
(d) none of these
Ans. (b)
x
Solution: The function f (x) = 3x + 4 is onto as for y Œ
Ê y-4ˆ
= y. The function f : R+ Æ R+, f (x) = x is
R, f Á
Ë 3 ˜¯
onto as for y Œ R+, f (y2) = y. f : R Æ R+, f (x) = x2 + 2 is not
onto as 1 ΠR+ has no pre-image.
Example 43: Which of the following functions is not
one-one
(a) f: R Æ R, f(x) = 2x + 5
(b) f: [0, p] Æ [–1, 1], f(x) = cos x
(c) f: [– p /2, p/2] Æ [1, 7] f (x) = 3 sin x + 4
(d) f : R Æ [–1, 1], f (x) = sin x
Ans. (d)
Solution: The function in (a) is one-one as 2x1 + 5 = 2x2
+ 5 is possible only if x1 = x2. The function in (b) is one-one
x - x2
as cos x1 = cos x2 if and only if sin 1
= 0 i.e., x1 = x2.
2
Similarly the function in (c) is also one-one. The function
in (d) is not one-one as f (p) = f (2p) = 0.
Example 44: Which of the following functions is nonperiodic
(a) f(x) = x – [x]
Ï1 if x is a rational number
(b) f(x) = Ì
Ó0 if x is an irrational number
8
8
+
1 + cos x 1 - cos x
(d) none of these
Ans. (d)
(c) f(x) =
1.16
Complete Mathematics—JEE Main
Solution: The function in (a) is periodic with period 1
and the function in (b) is also periodic since f (x + r) = f (x)
4
for every rational r. The function in (c) is equal to
sin x
and thus has period p.
ax
, x π – 1. Then, for what
Example 45: Let f (x) =
x +1
value of a is f (f (x)) = x ?
(b) - 2
(d) –1
(a) 2
(c) 1
Ans. (d)
Solution: f (f (x)) =
a f ( x)
a2x
=
f ( x ) + 1 (a + 1) x + 1
Thus f (f (x)) = x ¤ a2x = (a + 1)x2 + x
¤ (a2 – 1)x = (a + 1) x2
¤ (a + 1) ((a – 1) – (a + 1)x) = 0
Since (a – 1) – (a + 1) x π 0 for all x so a = – 1.
Example 46: Let R = {(x, y): x, y Œ R, x2 + y2 £ 25}
4
R¢ = ( x, y) : x, y Œ R, y ≥ x 2 then
9
(a) dom R « R¢ = [– 4, 4]
(b) range R « R¢ = [0, 4]
(c) range R « R¢ = [0, 5]
(d) R « R¢ defines a function.
Ans. (c)
{
}
Solution: The equation x2 + y2 = 25 represents a circle
4 2
with centre (0, 0) and radius 5 and the equation y =
x
9
represents a parabola with vertex (0, 0) and focus (0, 1/9).
Hence R « R¢ is the set of points indicated in the Fig.1.27
(-3, 4)
(0, 5)
(3, 4)
(3, 0)
Fig. 1.27
= {(x, y ): – 3 £ x £ 3,0 £ y £ 3]}. Thus
dom R « R ¢ = [-3, 3] and range R « R¢ = [0, 5] … [0, 4]
Since (0, 0) Œ R « R¢ and (0, 5) Œ R « R¢
\ 0 is related to 0 as well as 5.
Hence R « R¢ doesn’t define a function.
Example 47: In a factory 70% of the workers like
oranges and 64% like apples. If x% like both oranges and
apples, then
(a) x £ 34
(c) 34 £ x £ 64
Ans. (c)
(b) x ≥ 64
(d) none of these.
Solution: Let the total number of workers be 100. A, the
set of workers who like oranges and B, the set of workers
who like apples.
So
n(A) = 70, n(B) = 64, n ( A « B ) = x.
n ( A » B ) £ 100.
Also
fi
n(A) + n(B) – n ( A « B ) £ 100
fi
70 + 64 – x £ 100 fi x ≥ 34
n ( A « B ) £ n(B) fi x £ 64
Since
Hence
34 £ x £ 64.
Example 48: The Cartesian product of A ¥ A has 16
elements. S = {(a, b) Œ A ¥ A Áa < b}. (– 1, 2) and (0, 1)
are two elements belonging to S. The remaining elements of
S are given by.
(a) {(–1, 0), (– 1, 1), (0, 2), (1, 2)}
(b) {(–1, 0), (1, 1), (2, –1), (1, 2)}
(c) {(0, –1), (1, – 1), (0, 2), (1, 2)}
(d) none of these
Ans. (a)
Solution: ( -1, 2 ) Œ A ¥ A
fi
–1 Œ A, 2 Œ A and (0, 1) Œ A ¥ A fi 0 Œ A, 1 Œ A
So,
A = {– 1, 0, 1, 2} as A has four elements
and
S = {(– 1, 0 ), (–1, 1), (–1, 2), (0, 1), (0, 2), (1, 2)}.
Hence the required element of S are given by (a)
Example 49: If R and R¢ are two symmetric relations
(not disjoint) on a set A, then the relation R « R¢ is
(a) reflexive
(b) symmetric
(c) transitive
(d) none of these.
Ans. (b)
Let
fi
fi
fi
Solution:
(a, b) Œ R « R¢ for some a, b Œ A.
(a, b) Œ R and (a, b) Œ R¢
(b, a) Œ R and (b, a) Œ R¢ as R and R¢ are symmetric
(b, a) Œ R « R¢ showing that R « R¢ is symmetric.
Example 50: a, b, g denote respectively the sets
containing the letters in the names Apoorv, Mannan and
Manvi of three children playing together. Which of the
following is not correct.
(a) n (a « g ) Án (a » b » g )
(b) n (b « g ) Án (a » b » g )
(c) n (a » b » g ) = 8
(d) n (a » b » g ) = n (a » g )
Ans. (b)
a = {a, o, p, r, v}, b = {a, m, n}
g = {a, i, m, n, v}
a « b = {a}, a « g = {a, v}, b « g = {a, m, n}
a » b »g = {a, i, m, n, o, p, r, v} = a » g.
Solution:
Sets, Relations and Functions 1.17
Example 51: If f : R Æ R, defined by f (x) = x3 + 7, then
the value of f –1(71) and f –1 (–1) respectively are
(a) {4}, f
(b) f, {– 2}
(c) {4}, {–2}
(d) {2}, {– 4}
Ans (c)
Solution: f (x) = x3 + 7 = 71 fi x3 = 64 fi x = 4
fi
f –1 (71) = {4}
and f(x) = x3 + 7 = –1 fi x3 = – 8 fi x = –2 fi f –1(–1) = {–2}
Example 52: If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, B = {2,
4, 6, 8, 10, 12, 14, 16, 18} and N, the set of natural numbers
is the universal set, then A¢ » [(A » B) « B¢] is
(a) A
(b) A¢
(c) B
(d) N.
Ans. (d)
Solution: (A » B) « B¢ = A as A « B = f
fi
A¢ » [(A » B) « B¢] = A¢ » A = N.
Example 53: If X = {1, 2, 3, 4,}, then a one-one onto
mapping f: X Æ X such that f(1) = 1, f(2) π 2 and f(4) π 4 is
given by
(a) {(1, 1), (2, 3), (3, 4), (4, 2)}
(b) {(1, 1), (2, 4), (3, 1), (4, 2)}
(c) {(1, 2), (2, 4), (3, 2), (4, 3)}
(d) none of these
Ans. (a)
Solution: f in (a) is clearly one-one and onto also satisfies f(1) = 1, f(2) = 3 π 2, f(4) = 2 π 4.
2
Example 54: Let f(x) = (x + 1) – 1, x ≥ –1 then the set
{x : f(x) = f –1(x)} is equal to
(a) {0, – 1}
(b) {0, 1}
(c) {– 1, 1}
(d) {0}
Ans. (a)
Solution: Let y = (x + 1)2 – 1, x ≥ – 1
fi
(x + 1)2 = y + 1
fi
x+1=
fi
Thus
So
fi
fi
fi
y + 1 as x ≥ – 1
x=–1+
f –1(x) = – 1 +
y +1 , y ≥ – 1
x +1
–1
f (x) = f (x)
(x + 1) – 1 = – 1 + x + 1
2
x + 1 = 0 or (x + 1)3/2 = 1
x = – 1 or x = 0
Example 55: Let P = {q : sin q – cos q = 2 cos q} and
Q = {q : sin q + cos q = 2 sin q } be two sets, then
(a) P Ã Q and Q ~ P π f (b) Q À P
(c) P À Q
(d) P = Q.
Ans. (d)
Solution: sin q – cos q =
2 cos q
¤
sinq = ( 2 + 1 ) cos q
¤
cosq = ( 2 - 1 ) sin q
¤
fi
sinq + cosq =
P = Q.
2 sin q
Example 56: If A, B, C are three sets such that A « B =
A « C and A » B = A » C, then
(a) B = C
(b) A « B = f
(c) A = B
(d) A = C
Ans. (a)
Solution: Let x Œ B fi x Œ A » B
fi
x Œ A » C fi x Œ A or x Œ C
If x Œ A, then x Œ A « B = A « C fi x Œ C
So
x ΠB fi x ΠC.
Similarly x ΠC fi x ΠB, Hence B = C.
Example 57: The domain of the function
sin -1 ( x - 3)
f(x) =
is
9 - x2
(a) [1, 2]
(c) [1, 3]
Ans. (b)
(b) [2, 3]
(d) [1, 2]
Solution: x2 < 9 fi –3 < x < 3
and –1 £ x – 3 £ 1 fi 2 £ x < 3
Example 58: Let X = {1, 2, 3, 4, 5}. The number of
different ordered pairs (Y, Z) that can be formed such that
Y Õ X, Z Õ X, and Y « Z is empty is
(b) 53
(a) 25
2
(c) 5
(d) 35
Ans. (d)
Solution: For each x ΠX, we have three choices
x Œ Y, x œ Z; x œ Y, x Œ Z; x œ Y, x œ Z
So the required number of ordered pairs is 35.
Example 59: Let X = {1, 2, 3, 4}. The number of
equivalence relations that can be defined on X is
(a) 10
(b) 15
(c) 16
(d) 8
Ans. (b)
Solution: The number of equivalence relations
k -1
Bk =
 ( kn-1 )Bn ;
B0 = 1, B1 = 1, B2 = 2, B3 = 5
n=0
B4 =
( 30 ) B0 + ( 13 ) B1 + ( 32 ) B2 + ( 33 ) B3
= 1 + 3 + 3 ¥ 2 + 1 ¥ 5 = 15
Example 60: The function f(x) = sin
px
px
- cos
is
(n + 1)!
n!
1.18
Complete Mathematics—JEE Main
(a)
(b)
(c)
(d)
non periodic
periodic with period 2(n!)
periodic with period 2 (n + 1)!
periodic with period n!
f(x – 1) + f(x + 1) =
f(x – 2) + f(x) =
Solution: Since the period of sin x is 2p so the period
px
px
2p n!
of sin
is
= 2(n!). The period of cos
is
(
+ 1)!
n
n!
p
2(n + 1)! The period of f(x) = l.c.m. (2(n!), 2(n + 1)!) =
2(n + 1)!
Ê1- xˆ
= x 3 , x π –1, 1 and
Example 61: If f (x) f Á
Ë 1 + x ˜¯
f(x) π 0, then {f(–2)} (the fractional part of f(–2)) is equal to
2
(b) 1 3
(c) 1 2
(d) 0
... (i)
2 f ( x - 1)
... (ii)
2 f ( x + 1)
... (iii)
Replace x by x –1 in (i)
Ans. (c)
(a) 2 3
2 f ( x)
Replace x by x +1 in (ii)
f(x) + f(x + 2) =
Adding (ii) and (iii), we have
2 ( f(x – 1) + f(x + 1))
f(x – 2) + f(x + 2) + 2 f(x) =
=
2 f(x) = 2f(x)
2
fi
f(x –2) = – f (x + 2)
Replacing x by x + 2, we have
f(x) = –f(x + 4) = – ( f (x + 8)) = f(x + 8)
10
 f (3 + 8r )
= 11 ¥ f(3) = 11 ¥ 5 = 55.
r =0
Ans. (a)
Solution: Replacing x by
Example 63:
1- x
in the equation
1+ x
1- xˆ
3
( f ( x ))2 f ÊÁË
˜ =x
1+ x¯
sin 2 q
...(i)
Let f(q) =
We have
2
Ê Ê1- xˆˆ
Ê1- xˆ
ÁË f ÁË
˜¯ ˜¯ f ( x ) = ÁË
˜
1+ x¯
1+ x
3
... (ii)
Solving (i) and (ii), we have
2
1- xˆ3
Ê x3 ˆ
= ÊÁ
f
x
(
)
ÁË ( f ( x ))2 ˜¯
Ë 1 + x ˜¯
( f ( x ))3
fi
x6
Ê1+ xˆ
= Á
Ë 1 - x ˜¯
then f is
(a)
(b)
(c)
(d)
Ans. (c)
2
sin q
1 + cos q
4 sin 4q
1 + sin 2 q
cos2 q
4 sin 4q
3
Solution: f(q) =
-1
-1
0
1
1
0
1 + sin 2 q
cos2 q
4 sin 2q
(R1 Æ R1 - R3
4
Ê -1ˆ
f(–2) = 4 Á ˜ = Ë 3¯
3
{f(–2)} = -
R2 Æ R2 - R3 )
4 È 4˘
4
2
- Í- ˙ = - + 2 = .
3 Î 3˚
3
3
Example 62: Let f(x) be a function such that f(x – 1) +
10
f(x + 1) =
2 f ( x ) for all x ΠR. If f(3) = 5 then
is equal to
(a) 50
(c) 0
Â
r =0
(b) 55
(d) 10
Ans. (b)
Solution: The given equation is
1 + 4 sin 4q
2
a non periodic function
periodic with period p
periodic with period p 2
odd function
Ê1+ xˆ
f(x) = x 2 Á
Ë 1 - x ˜¯
fi
cos2 q
f (3 + 8 r )
=
0
-1
2
1 + sin q
+ 4 sin 4q
0
1
1
0
cos2 q
4 sin 4q
(C1 Æ C1 + C3)
= – (cos2q + 1 + sin2q + 4 sin 4 q)
= – 2 (1 + 2 sin 4 q)
2p p
= .
which is periodic function with period
4
2
Sets, Relations and Functions 1.19
Assertion-Reason Type Questions
Example 64: Let R be the real line. Consider the
following subsets of the plane R ¥ R:
S = {(x, y) : y = x + 1 and 0 < x < 2}
T = {(x, y) : x – y is an integer}
Statement-1 : T is an equivalence relation on R but S is not
an equivalence relation on R.
Statement-2 : S is neither reflexive nor symmetric but T is
reflexive, symmetric and transitive.
Ans. (a)
Solution: Since x π x + 1, (x, x) œ S, so S is not reflexive.
Next x, y Œ S, fi y = x + 1 fi x = y – 1 fi (y, x) œ S, so is
not symmetric.
Since x Рx = 0 is an integer (x, x) ΠT " x ΠT
fi
T is reflexive.
Again (x, y) ΠT fi x Рy is an integer
fi
y Рx is also an integer fi (y, x) ΠT
So T is symmetric.
Also (x, y) ΠT, (y, z) ΠT.
fi
x – y and y – z are integers
fi
x – z = (x – y) – (y – z) is also an integer
fi
(x, z) ΠT
So T is Transitive.
Which shows that statement-2 is true and hence statement-1
is also true.
Example 65: Consider the following relations.
R = {(x, y) | x, y are real numbers and x = wy for some
rational number w}
ÏÊ m p ˆ ¸
S = ÌÁ , ˜ ˝ m, n, p, q, are integer such that n◊q π 0
ÓË n q ¯ ˛
and qm = pn}
Statement-1: S is an equivalence relation but R is not an
equivalence relation.
Statement-2: R and S both are symmetric.
Ans. (c)
Solution: Since (0, 1) Œ R but (1, 0) œ R, R is not symmetric and hence is not an equivalence relation so statement-2 is false.
m p
Next, For the relation S, qm = pn fi
=
n q
m p
Ê m pˆ
Thus Á , ˜ Œ S fi
which shows that S is
=
Ë n q¯
n q
reflexive and symmetric
Ê m pˆ
Ê p rˆ
Again, Á , ˜ ŒS and Á , ˜ Œ S
Ë n q¯
Ë q s¯
m p r
m r
= =
fi ÊÁ , Œ S ˆ˜
Ën s
¯
n q s
Thus S is transitive and hence S is an equivalence relation.
So, statement 1 is true.
fi
Example 66: Let R be a relation on the set N of natural
numbers defined by n Rm ¤ n is a factor of m (i.e. n | m).
Statement-1: R is not an equivalence relation
Statement-2: R is not symmetric
Ans. (a)
Solution: Statement-2 is true as 2 | 6 fi 2R6 but 6 does
not divide 2 so R is not symmetric fi R is not an equivalence
relation and the statement-1 is also true.
Example 67: Let A = {1, 2, 3} and B = {3, 8}
Statement-1: (A » B) ¥ (A « B) = {(1, 3), (2, 3), (3, 3),
(8, 3)}
Statement-2: (A ¥ B) « (B ¥ A) = {(3, 3)}
Ans. (b)
Solution: A » B = {1, 2, 3, 8}, A « B = {3}
(A » B) ¥ (A « B) = {(1, 3), (2, 3), (3, 3), (8, 3)}
Statement-1 is True.
(x, y) Œ (A ¥ B) « (B ¥ A)
fi
(x, y) Œ A ¥ B and (x, y) Œ B ¥ A
fi
x Œ A « B, y Œ A « B
fi
{(3, 3)} = (A ¥ B) « (B ¥ A) fi Statement-2 is
also true but is not a correct explanation for statement-1.
fi
fi
Example 68: Statement-1: The number of bijective
functions from the set A containing 100 elements to itself
is 2100.
Statement-2: The total number of bijections from a set containing n elements to itself is n!
Ans. (d)
Solution: Statement-2 is true and so, statement-1 is False.
Example 69: Statement-1: f : R Æ R is a function
x-3
defined by f(x) = 5x + 3. If g = f –1, then g(x) =
.
5
Statement-2: If f : A Æ B is a bijection and g : B Æ A is the
inverse of f, then fog is the identity function on A.
Ans. (c)
y-3
Solution: Let y = 5x + 3 fi x =
.
5
x-3
fi
is the inverse of f, so statement-1 is True.
g(x) =
5
1.20
Complete Mathematics—JEE Main
Statement-2 is false because g : B Æ A and f : A Æ B
fi
fog : B Æ B and g = f –1 fi fog is an identity
function on B.
Example 70: Let X and Y be two sets.
Statement-1: X « (Y » X)¢ = f
Statement-2: If X » Y has m elements and X « Y has n
elements then symmetric difference X D Y
has m – n elements
Ans. (b)
Solution: X « (Y » X)¢ = X « (Y ¢ « X ¢) = X « X ¢ « Y ¢ = f.
fi
Statement-1 is True.
X D Y = (X ~ Y) » (Y ~ X) = (X » Y) ~ (X « Y)
fi
number of elements in X D Y = m – n.
fi
Statement-2 is True but does explain statement-1.
Example 71: Let f be a function defined by
f(x) = (x – 1)2 + 1, (x ≥ 1)
Statement-1: The set {x : f(x) = f –1(x)} = {1, 2}
Statement-2: f is a bijection and f –1(x) = 1 + x - 1 , x ≥ 1
Ans. (a)
Solution: Let y = f(x) = (x – 1)2 + 1
fi
y – 1 = (x – 1)2 fi x = 1 + y - 1 , y ≥ 1
Thus f –1(x) = 1 +
x - 1 , x ≥ 1. So statement-2 is true.
–1
Now f(x) = f (x)
fi
(x – 1)2 =
x -1
fi
x - 1 [( x - 1)3 / 2 - 1] = 0
x = 1, 2.
fi
So statement-1 is true and statement-2 is a correct explanation for statement-1.
Example 72: Let R be the set of real numbers
Statement-1: A = {(x, y) Œ R ¥ R : y + x is an integer} is an
equivalence relation on R.
Statment-2: B = {(x, y) Œ R ¥ R : y = a x for some rational
number a} is not equivalence relation on R.
Ans. (d)
Solution: A is neither reflexive nor transitive as x + x
may not be integer " x ŒR and if x + y and y + z are integers,
x + z may not be an integer for x, y, z ŒR.
So statement-1 is false.
Statememt-2 is true as B is not symmetric, because ( 3 , 0)
ΠB as 0 =
3 ¥ 0 for a = 0, but (0,
3 ) œ B.
Example 73: Consider the following relation R on the
set of real square matrices of order 3.
R = { (A, B): A = P –1BP for some invertible matrix P}
Statement-1: R is an equivalence relation.
Statement-2: For any two invertible 3 ¥ 3 matrices M and
N, (MN)–1 = N –1M –1.
Ans. (b)
Solution: Statement-2 in true (See Text.)
In statement-1, A = I–1 A I
for all real square matrices A of order 3.
fi (A, A) ΠR fi R is reflexive,
Next, let (A, B) ΠR
fi $ a invertible matrix P of order 3.
such that A = P –1 B P
fi B = P A P–1 = (P –1)–1A (P –1)
fi R in symmetric
If Now (A, B) ΠR and (B, C) ΠR
Then $ invertible matrices P and Q
of order 3 such that
A = P –1 B P and B = Q –1 C Q.
fi A = P –1 Q –1 C Q P = (QP)–1 C QP (From
statement-2)
fi (A, C) ΠR and thus R in transitive. Hence R is an equivalence relation and the statement-1 in also true but statement-2 is not a correct explanation for it.
LEVEL 2
Straight Objective Type Questions
Example 74: From 50 students taking examinations in
Mathematics, Physics and Chemistry, 37 passed Mathematics,
24 Physics and 43 Chemistry. At most 19 passed Mathematics
and Physics, at most 29 Mathematics and Chemistry and at
most 20 Physics and Chemistry. The largest possible number
that could have passed all three exams is
(a) 10
(b) 12
(c) 9
(d) none of these.
Ans. (d)
Solution: The given conditions can be expressed as
n (M » P » C) = 50, n (M) = 37, n(P) = 24,
n(C) = 43, n(M « P) £ 19,
n(M « C) £ 29 and n(P « C) £ 20,
n (M » P » C) = n(M) + n(P) + n(C) - n(M « P)
- n(M « C) - n (P « C) + n(M
« P « C)
fi
50 = 37 + 24 + 43 - n (M « P) - n(P « C)
Sets, Relations and Functions 1.21
- n(M « C) + n(M « P « C)
fi
n (M « P « C) £ n (M « P) + n (M «C)
+ n (P « C) - 54.
Therefore, the number of students that could have
passed all three exams is at most
19 + 29 + 20 - 54 = 14.
Example 75: Suppose A1, A2, º, A30 are thirty sets each
having 5 elements and B1, B2 º, Bn are n sets each with 3
30
elements, let
∪
n
Ai =
i =1
∪
Bj = S and each element of S
j =1
belongs to exactly 10 of the Ai’s and exactly 9 of the Bj’s.
Then n is equal to
(a) 15
(b) 3
(c) 45
(d) none of these
Ans. (c)
30
1
Solution: S = ∪ Ai, so n(S) =
(5 ¥ 30) = 15 (since
10
i =1
element in the union S belongs to exactly 10 of the Ai’s).
n
Again S = ∪ Bi so
i =1
n (S) = 1/9(3 ¥ n) = n/3 = 15 fi n = 45
Example 76: Let R = {(3, 3), (6, 6), (9, 9) (12, 12), (6, 12),
(3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9,
12}.
The relation is
(a) an equivalence relation.
(b) reflexive and symmetric only.
(c) reflexive and transitive only
(d) reflexive only.
Ans. (c)
Solution: R is reflexive as
(3, 3), (6, 6), (9, 9), (12, 12) ΠR.
R is not symmetric as (6, 12) Œ R but (12, 6) œ R.
R is transitive as the only pair which needs verification is (3, 6) and (6, 12) ΠR fi (3, 12) ΠR.
Example 77: Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)}
be a relation on the set A = {1, 2, 3, 4}, the relation R is
(a) not symmetric
(b) transitive
(c) a function
(d) reflexive
Ans. (a)
Solution: R is not symmetric as (2, 3) ΠR but
(3, 2) œ R. It is not transitive as (1, 3), (3, 1) Œ R. but (1, 1)
œ R so is not reflexive also.
Again as (2, 4) and (2, 3) ΠR, it is not a function.
Example 78: Let I be the set of integers, N the set of
non-negative integers; Np the set of non-positive integers ;
E is the set of even integers and P is set of prime numbers.
Then
(a) N « Np = f
(c) N D Np = I – {0}
Ans. (c)
(b) I – N = Np
(d) E « P = f.
Solution: N « Np = {0},
I – N = {..., – 2, – 1}
N D Np = {N – Np) » (Np – N)
= {1, 2, 3 - -} » { ..., – 3, – 2, – 1}
= I ~ {0}
and
E « P = {2}.
Example 79: If n(A) = n then n{(x, y, z); x, y, z ΠA,
x π y, y π z, z π x}=
(b) n(n – 1)2
(a) n3
2
(c) n (n – 2)
(d) n3 – 3n2 + 2n
Ans. (d)
Solution: There are n choices for the first coordinate,
n – 1 choices for second coordinate and n – 2 choices for the
third coordinate, hence n ({(x,y, z); x, y, z Œ A, x π y π z}) =
n(n – 1) (n – 2) = n3 – 3n2 + 2n.
Example 80: If A = {x : x Œ I, – 2 £ x £ 2},
B = {x Œ I, 0 £ x £ 3}, C = {x: x Œ N, 1 £ x £ 2} and
D = {x, y) Œ N × N; x + y = 8}. Then
(a) n(A » (B » C) = 5 (b) n(D) = 6
(c) n(B » C) = 5
(d) none of these
Ans. (d)
Solution: A = {– 2, – 1, 0, 1, 2}, B = {0, 1, 2, 3},
C = {1, 2}
so
B » C = {0, 1, 2, 3},
A » (B » C) = {–2, –1, 0, 1, 2, 3}
so n(A » (B » C) = 6, n (B » C) = 4
and
D = {(1, 7), (2, 6), (3, 5), (4, 4), (5,3),
(6, 2), (7, 1)} so
n(D) = 7.
Example 81: If A = {4n Р3n Р1| n ΠN} and
B = {9n – 9: n Œ N}, then A»B is equal to
(a) B
(b) A
(c) N
(d) none of these
Ans. (a)
Solution: It can be shown by induction that 9 Á4n – 3n – 1
for every n Œ N. Thus A Õ B. Clearly A π B as 27 Œ B but
27 œ A. Thus A » B = B.
Example 82: A and B are two sets having 3 and 4 elements respectively and having 2 elements in common. The
number of relations which can be defined from A to B is
(a) 25
(b) 210 – 1
12
(c) 2
(d) none of these.
Ans. (c)
Solution: The number of elements in A ¥ B is 12. Hence
the number of subsets of A ¥ B is 212, which includes the
empty set
1.22
Complete Mathematics—JEE Main
Example 83: If R and S are two symmetric relations then
(a) R o S is a symmetric relation
(b) S o R is a symmetric relation
(c) R o S –1 is a symmetric relation
(d) R o S is a symmetric relation if and only if RoS =
SoR.
Ans. (d)
Solution: Since R and S are symmetric relations so
R –1 = R and S –1 = S. But (R o S)–1 = S –1 o R –1 = S o R. Thus
RoS is symmetric if and only if RoS = SoR.
Example 84: Let A be the set of all determinants of order
3 with entries 0 or 1 only, B the subset of A consisting of all
determinants with value 1, and C the subset consisting of all
determinants with value –1. Then if n(B) and n(C) denote the
number of elements in B and C, respectively, we have
(a) C = f
(b) n(B) = n(C)
(c) A = B » C
(d) n(B) = 2n(C)
Ans. (b)
Solution: C cannot be the empty set because, for instance,
0 1 0
1 0 1
Р1 = 1 0 0 ΠC. We also have 1 1 0 = 2,
0 0 1
0 11
so A π B » C. In general, the determinant
a11 a12 a13
a21 a22 a23 =
a31 a32 a33
a11a22 a33 + a12 a23 a31 + a13 a21a32
- a11a22 a32 - a12 a21a33 - a13 a31a22 ,
with the a’s being 0 or 1, equals 1 only if a11a22a33 = 1 and
the remaining terms are zero; if a12 a23 a31 = 1 and the remaining terms are zero; or if a13 a21 a32 = 1 and the remaining terms are zero. Since there are three similar relations
for determinants that equal - 1, we must have n (B) = n (C).
Example 85: The domain of the function
1 ˆ ˆ
Ê
Ê
f(x) = log2 Á - log1 / 2 Á 1 + 4 ˜ - 1˜ is
Ë
Ë
x¯ ¯
(a) 0 < x < 1
(c) x ≥ 1
Ans. (a)
(b) 0 < x £ 1
(d) x > 1
Solution: For f to be defined we must have log1/2
1
1 ˆ
Ê
–1 –1
ÁË 1 + 4 ˜¯ < -1 ¤ 1 + 4 > (2 ) = 2 which is possible
x
x
1
if and only if 4 > 1 i.e., 0 < x < 1.
x
Hence the domain of the given function is {x : 0 < x < 1}.
Example 86: The domain of definition of
f(x) =
1
Ê x - 1ˆ
log0. 4 Á
¥ 2
is
Ë x + 5 ˜¯
x - 36
(a)
(b)
(c)
(d)
Ans. (c)
{x : x < 0, x π – 6}
{x : x > 0, x π 1, x π 6}
{x : x > 1, x π 6}
{x : x ≥ 1, x π 6}
Ê x -1 ˆ
log0 ◊ 4 Á
to be defined, we must
Ë x + 5 ˜¯
1
x -1
have 0<
< 1, which is true if x > 1. Morever, 2
x - 36
x+5
is defined for x π ± 6. Hence the domain of f is
{x : x > 1, x π 6}.
Solution: For
Example 87: The set of all x for which f(x) = log x - 2 2
x+3
1
are both not defined is
and g(x) =
x2 - 9
(a) (– 3, 2)
(b) [– 3, 2)
(c) (– 3, 2]
(d) [–3, 2]
Ans. (d)
Solution: The function f is not defined for - 3 £ x £
2 and g is not defined for those x for which x2 - 9 £ 0 i.e.,
x Π[Р3, 3]. Thus f and g are not defined on [-3, 2].
Example 88: If f (x) is a polynomial satisfying f (x)
f (1/x) = f (x) + f (1/x) and f (3) = 28, then f (4) is given by
(a) 63
(b) 65
(c) 67
(d) 68
Ans. (b)
Solution: Any polynomial satisfying the functional
equation f (x) ◊ f (1/x) = f (x) + f (1/x) is of the form xn + 1 or
- xn + 1. If 28 = f (3) = - 3n + 1 then 3n = - 27 which is not
possible for any n. Hence 28 = f (3) = 3n + 1 fi 3n = 27 fi n
= 3. Thus f (x) = x3 + 1, so f (4) = 43 + 1 = 65.
Example 89: Part of the domain of the function
cos x - 1 / 2
lying in the interval [–1, 6] is
6 + 35 x - 6 x 2
(a) [–1/6, p/3] » [5p/3, 6]
(b) (–1/6, p/3] » [5p/3, 6)
(c) (–1/6, 6)
(d) none of these
Ans. (a)
f (x) =
Solution: The function f is meaningful only if cos x - 1/2
≥ 0, 6 + 35 x - 6x2 > 0 or cos x - 1/2 £ 0, 6 + 35x - 6x2< 0
i.e., cos x ≥ 1/2, (6 - x) (1 + 6x) > 0 or cos x £ 1/2, (6 - x)
(1 + 6x) < 0. These inequalities are satisfied if
x Œ (-1/6, p/3] » [5p/3, 6).
Example 90: Let f: R Æ R be a function defined by
e| x| - e- x
f (x) = x
. Then
e + e- x
(a) f is both one-one and onto
(b) f is one-one but not onto
Sets, Relations and Functions 1.23
(a) f(A « B) = f(A) « f(B) (b) f(A « B) = [0, 1]
(c) f is onto but not one-one
(d) f is neither one-one nor onto
Ans. (d)
Solution: f is not one-one as f (0) = 0 and f (-1) = 0.
f is also not onto as for y = 1 there is no x ΠR such that
f (x) = 1. If there is such a x Œ R then e|x| – e–x = ex + e–x,
clearly x π 0. For x > 0, this equation gives - e–x = e–x which
is not possible. For x < 0, the above equation gives ex = - e–x
which is also not possible.
Example 91: Let f (x) = x2 and g(x) = 2x then the solution
set of fog (x) = g o f (x) is
(a) R
(b) {0}
(c) {0, 2}
(d) none of these
Ans. (c)
x
x 2
2x
Solution: fog (x) = f (g (x)) = f (2 ) = (2 ) = 2 and gof
x2
2
x2
2x
(x) = g (f (x)) = g (x ) = 2 . Thus the solution of 2 = 2 is
given by x2 = 2x which is x = 0, 2.
Example 92: A function f : R Æ R satisfies the equation
f (x) f (y) Рf (xy) = x + y for all x, y ΠR and f (1) > 0, then
(a) f (x) = x + 1/2
(b) f (x) = (1/2)x + 1
(c) f (x) = (1/2)x – 1
(d) f(x) = x + 1
Ans. (d)
Solution: Taking x = y = 1, we get
f (1) f (1) – f (1) = 1 + 1 fi f (1)2 – f (1) – 2 = 0
fi (f (1) – 2) (f (1) + 1) = 0 fi f (1) = 2 (∵ f (1) > 0)
Taking y = 1, we get
f (x) f (1) – f (x) = x + 1 fi 2f (x) – f (x) = x + 1
fi
f (x) = x + 1.
Example 93: If f : [1, •) Æ [2, •) is given by f (x) = x +
1/x then f –1(x) equals
(a)
(c)
Ans. (a)
x + x2 - 4
2
(b)
x - x2 - 4
2
2
(d) 1 + x - 4
x
1 + x2
Solution: y = x + 1/x fi x2 – xy + 1 = 0
x=
y ± y2 - 4
2
x=
y + y2 - 4
.
2
f – 1(x) =
x + x2 - 4
.
2
fi
Since x Œ [1, •)
Hence
so
Example 94: Consider the function f = {(x, sin x) | – • <
È p˘
È p˘
x < •}. Let A = Í0, ˙ and B = Í0, ˙ then
Î 6˚
Î 2˚
(c) f(A) « f(B) = [0, 1]
Ans. (a)
È 1˘
(d) f(A) » f(B) = Í0, ˙
Î 2˚
Solution: f(x) = sin x is an increasing function on
1
p
[0, p / 2] so f(A) = ÈÍ0, sin ˘˙ = ÈÍ0, ˙˘ and f(B) = [0, 1].
6˚
Î 2˚
Î
1
È
È p˘
˘
Thus f(A) « f(B) = Í0, ˙ . Also A « B = Í0, ˙ , so f(A « B)
Î 2˚
Î 6˚
1
p
È 1˘
È
˘
˘ È
= Í0, sin ˙ = Í0, ˙ . Thus f(A) « f(B) = Í0, ˙ = f(A « B),
6 ˚ Î 2˚
Î 2˚
Î
Also f(A) » f(B) = [0, 1].
Example 95: Let f(x) be a polynomial of even degree
Ê
Ê 1 ˆˆ
satisfying f(2x) Á 1 - f Á ˜ ˜ + f 16 x 2 y = f(–2) – f(4xy)
Ë
Ë 2x ¯ ¯
for all x, y Œ R ~ {0} and f(4) = –255, f(0) = 1. Then the
f (2 ) + 1
is
value of
2
(a) 4
(b) 5
(c) 7
(d) 6
Ans. (c)
1
Solution: Replacing y by
in the given functional
8x2
equation, we obtain
Ê
Ê 1 ˆˆ
Ê 1ˆ
f (2x) Á 1 - f Á ˜ ˜ + f (2) = f (–2) – f Á ˜ .
Ë
Ë 2x ¯ ¯
Ë 2x ¯
(
)
Since f is an even function so f (2) = f (–2),
Ê 1ˆ
Ê 1ˆ
so
f (2 x ) - f (2 x ) f Á ˜ = - f Á ˜
Ë 2x ¯
Ë 2x ¯
Ê 1ˆ
Ê 1ˆ
fi
f (2 x ) + f Á ˜ = f (2 x ) f Á ˜
Ë 2x ¯
Ë 2x ¯
x
Replacing x by , we have
2
Ê 1ˆ
Ê 1ˆ
f ( x) + f Á ˜ = f ( x) f Á ˜
Ë x¯
Ë x¯
n
Since f is a polynomial, so f (x) = ± x + 1 (see Example 88)
But –255 = f(4) = ± 4n + 1. Only negative sign is possible, thus
4n = 256 fi n = 4. i.e. f (x) = –x4 + 1. f (2) = –24 + 1 = –15
f (2 ) + 1
14
= = 7.
2
2
1
Example 96: The domain of the function f(x) =
sin x
x-5
+ 3 sin x + log10 2
is
x - 10 x + 24
(a) {(2 k p, (2k + 1)p) : k ΠI}
(b) {(2 k p, (2k + 1)p ) : k ΠN}
(c) (6, •) » (4, 5)
(d) {(2 k p, (2 k + 1)p) : k Œ I} » (6, •)
Ans. (b)
1.24
Complete Mathematics—JEE Main
Solution: The domain of 3 sin x is R.
1
= {x : sin x > 0}
The domain of
sin x
= {(2kp, (2k + 1)p) : k ΠI}
The domain of log10
x-5
2
x - 10 x + 24
= log10
x-5
( x - 6)( x - 4)
= {x : x π 5, x > 5, x > 6, x > 4 or
x π 5, x < 5, 4 < x < 6}
= (4, 5) « (6, •)
Thus the domain of f (x)
= {(2 k p, (2k + 1)p) : k Œ I} « {(4, 5) » (6, •)}
= {(2 k p, (2k + 1)p) : k ΠN}
(since (4, 5) does not intersect {(2 k p, 2k + 1)p : k ΠI})
EXERCISE
Concept-based
Straight Objective Type Questions
1. Let U be a Universal set and n(U ) = 12. If A, B Õ U
are such that n(B) = 6 and n(A « B) = 2 then n(A » B¢)
is equal to
(a) 6
(b) 10
(c) 7
(d) 8
2. Let R be a relation on R defined as a R b if |a| £ b.
Then, relation R is
(a) reflexive
(b) symmetric
(c) transitive
(d) not antisymetric
3. Let f(x) = ax + b, x ΠR, and g(x) = x + d, x ΠR,
then fog = gof if and only if
(a) f(a) = g(c)
(b) f(d) = g(b)
(c) f(b) = g(d)
(d) f(c) = g(a)
sin x
is
4. The domain of the function f(x) =
|x|- x
(a) R
(c) R+
(b) R ~ {0}
(d) R–
(R+ is the set of positive real numbers and R – is the
set of negative real numbers)
5. The domain of y = cos–1 (1 – 2x) is
(a) [– 1, 1]
(b) [0, 1]
1 1
(c) [–1, 0]
(d) È- , ˘
ÍÎ 2 2 ˙˚
6. If f(x) = sin x – cos x is written as f1(x) + f2(x) where
f1(x) is even and f2(x) is odd then
(b) f1(x) = – cos x
(a) f1(x) = cos x
(c) f2(x) = – sin x + cos x (d) f2(x) = sin (2p – x)
7. The range of y = 1 – sin x is
(a) [–1, 1]
(b) [0, 1]
(c) [–1, 2]
(d) [0, 2]
1- x
is
8. The function f(x) = x2 log
1+ x
(a) a periodic function
(c) an odd function
(b) a bounded function
(d) an even function
LEVEL 1
Straight Objective Type Questions
9. If A = {2, 3, 4, 5, 7}, B = {1, 2, 4, 7, 9} then
((A ~ B) » (B ~ A)) « A is equal to
(a) {3, 5}
(b) {2, 4}
(c) {3, 7}
(d) {2, 7}
10. If A = {2x : x ΠN}, B = {3x : x ΠN} and
C = {5x : x Œ N) then A « (B « C) is equal to
(a) {15, 30, 45, ... }
(b) {10, 20, 30 ... }
(c) {30, 60, 90, ... }
(d) {7, 14, 21 ... }.
11. If X and Y are two sets such that
X « Y = X » Y, then
(a) X Ã Y, X π Y
(b) Y Ã X, Y π X
(d) none of these
(c) X = Y
12. If X, Y and A are three sets such that A « X = A «
Y and A » X = A » Y then
(a) X Ã Y
(b) Y Ã X
(c) X = Y
(d) none of these
13. If A = {x : x ΠR and satisfy x2 Р15x + 56 = 0}
B= {x : x Œ N and x + 5 £ 14}
and C = {x : x ΠN and x/112}.
Then A » (B « C) is equal to
(a) {1, 2, 3,....8}
(b) {8, 7}
(c) {1, 2, 8, 7}
(d) {1, 2, 4, 7, 8}
Sets, Relations and Functions 1.25
14. If A is the set of letters needed to spell “MATHEMATICS” and B is the set of letters needed to spell
STATISTICS, then
(a) A Ã B
(b) A ~ B = f
(c) A D B = A ~ B
(d) none of these
15. If A and B are two sets such that n(A » B) = 36,
n(A « B) = 16 and n(A ~ B) = 15, then n(B) is equal to
(a) 21
(b) 31
(c) 20
(d) 52
16. The maximum number of sets obtainable from A and
B by applying union and difference operations is
(a) 5
(b) 6
(c) 7
(d) 8
17. If A and B both contain same number of elements and
are finite sets then
(a) n(A » B) = n(A « B) (b) n(A ~ B) = n(B ~ A)
(c) n(A D B) = n(B)
(d) n(A ~ B) = n(A)
18. If A D B = A » B then
(a) A = B
(c) A D B = f
(b) A « B = f
(d) A D B = A ~ B
19. In a class 60% passed their Physics examination and
58% passed in Mathematics. Atleast what percentage
of students passed both their Physics and Mathematics
examination?
(a) 18%
(b) 17%
(c) 16%
(d) 2%
20. If the relation R: A Æ B, where A = {1, 2, 3} and
B = {1, 3, 5} is defined by
R = {(x, y): x < y, x ΠA, y ΠB}, then
(a) R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5)}
(b) R = {(1, 1), (1, 5), (2, 3), (3, 5)}
(c) R–1 = {(3, 1), (5, 1), (3, 2), (5, 3)}
(d) R–1 = {(1, 1), (5, 1), (3, 2), (5, 3)}
21. The relation R defined on the set A = {1, 2, 3, 4, 5}
by
R = {(x, y): |x2 – y2| < 16}
is given by
(a) R1 = {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}
(b) R2 = {(2, 2), (3, 2), (4, 2), (2, 4)}
(c) R3 = {(3, 3), (4, 3), (5, 4), (3, 4)
(d) none of these
22. Let L be the set of all lines in a plane and R be
a relation on L defined by l1R l2 if and only if
l1 ^ l2 then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) an equivalence relation
23. For non-empty subsets A and B,
(a) Any subset of A × B defines a function from A to B.
(b) Any subset of A × B defines an equivalence relation
(c) Any subset of A × A defines a function on A
(d) none of these.
24. Let f (x) = (x + 1)2 – 1 (x ≥ –1). Then the set
S = {x : f (x) = f –1(x)} contains
Ï
-3 + i 3 -3 - i 3 ¸
,
(a) Ì0, - 1,
˝
2
2
Ó
˛
(b) {0, 1, –1}
(c) {0, –1}
(d) none of these
25. If a set A has n elements then the number of all relations on A is
2
2
(b) 2 n – 1
(a) 2 n
n
(c) 2
(d) none of these
26. The function f: R Æ R given by f (x) = 3 – 2 sin x
is
(b) onto
(a) one-one
(c) bijective
(d) none of these
27. Which of the following are functions?
(a) {(x, y): y2 = 4ax, x, y ΠR}
(b) {(x, y): y = |x|, x, y ΠR}
(c) {(x, y): x2 + y2 = 1, x, y ΠR}
(d) {(x, y): x2 Рy2 = 1, x, y ΠR}
28. If f: R Æ R defined by f (x) = x4 + 2 then the value
of f –1 (83) and f –1(– 2) respectively are
(a) f, {3, –3}
(b) {3, –3}, f
(c) {4, – 4}, f
(d) {4, – 4}, {2, –2}.
29. The minimum number of elements that must be added
to the relation R = {(1, 2), (2, 3)} on the sub set {1,2,3}
of natural numbers so that it is an equivalence relation
is
(a) 4
(b) 7
(c) 6
(d) 5
30. Let X be a non-empty set and P(X) be the set of all
subsets of X. For A, B ΠP(X), ARB if and only if
A « B = f then the relation
(a) R is reflexive
(b) R is symmetric
(c) R is transitive
(d) R is an equivalence relation
31. Let f be a function satisfying 2f (x) – 3f (1/x) = x2 for
any x π 0. Then the value of f (2) is
(a) – 2
(b) – 7/4
(c) – 7/8
(d) 4
1
x
( x + 1)
32. If f (x) =
2x
x( x - 1)
( x + 1) x
3 x( x - 1) x( x - 1)( x - 2) x( x - 1)( x + 1)
then
1.26
Complete Mathematics—JEE Main
f (50) + f (51) + …… + f (99) is equal to
(a) 0
(b) 1275
(c) 3725
(d) none of these
33. The domain of definition of the functions y(x) given
by the equation ax + ay = a (a > 1) is
(a) 0 < x £ 1
(b) 0 £ x £ 1
(c) – • < x < 1
(d) – • < x £ 0
x - [ x]
, where [x] denotes the greatest
1 + x - [ x]
integer less than or equal to x, then the range of f is
(a) [0, 1/2]
(b) [0, 1)
(c) [0, 1/2)
(d) [0, 1]
34. Let f (x) =
35. If 2f (x2) + 3 f (1/x2) = x2 – 1, then f (x2) is
(b) (1 – x2)/5x
(a) (1 – x4)/5x2
2
4
(c) 5x /(1 – x )
(d) none of these
36. If f (x + 3y, x – 3y) = 12xy, then f (x, y) is
(a) 2xy
(b) 2(x2 – y2)
2
2
(d) none of these
(c) x – y
37. If the function f : [1, •) Æ [1, •) is defined by
f (x) = 2x(x–1) then f –1(x) is
(a) (1/2)x (x –1)
(c)
1
1 - 1 + 4 log2 x
2
(
(b)
)
1
1 + 1 + 4 log2 x
2
(
)
(d) not defined
38. If n (A) = 3 and n (B) = 5 then number of one-one
functions that can be defined from A to B is
(a) 30
(b) 40
(c) 120
(d) 60
39. Let f: {x, y, z} Æ {1, 2, 3} be a one-one function. If
it is given that exactly one of the following statements
is true,
Statement-1: f(x) = 1, Statement-2: f(y) π 1, Statement-3: f(z) π 2.
then f –1 (1) is
(a) x
(b) y
(c) z
(d) none of these
40. The value of n Πz for which the function
f (x) =
(a) 2
(c) 4
sin nx
has 4p as its period is
sin ( x / n)
(b) 3
(d) 5
41. Let R be a relation on N defined by
R = {(m, n): m, n ΠN and m = n2}.
Which of the following is true.
(a) (n, n) ΠR . " n ΠN
(b) (m, n) ΠR fi (n, m) ΠR
(c) (m, n) ΠR, (n, p) ΠR fi (m, p) ΠR
(d) none of these
42. Of the number of three athletic teams in a school, 21
are in the basketball team, 26 in hockey team and 29
in the football team, 14 play hockey and basketball,
15 play hockey and football, 12 play football and
basketball and 8 play all the games. The total number
of members is
(a) 42
(b) 43
(c) 45
(d) none of these
43. E, I, R, O denote respectively the sets of all equilateral, isosceles, right angled and obtuse angled triangles
in a plane, then which of the following is not true.
(a) R « E = f, R « I π f
(b) E « O = f, O « I π f
(c) E « I = f, E « O π f
(d) E « I π f, EÃ I
44. If A = {3n : n Œ N, n £ 6}, B = {9n : n Œ N, n £ 4}
then which of the following is false
(a) A D B = {6561}
(b) A ~ B = {3, 27, 243}
(c) A « B = {9, 81, 729}
(d) A » B = {3, 9, 27, 81, 243, 729, 6561}
45. If f : R Æ R given by f(x) = ax + sin x + a, then f is
one-one and onto for all
(a) a ΠR
(b) a Œ R ~ [–1, 1]
(c) a ΠR ~ {0}
(d) a Œ R ~ {–1}
n
46. If f: (0, p) Æ R is given by f(x) =
 [1 + sin kx ] , [x]
k =1
denotes the greatest integer function, then the range
of f(x) is
(a) {n–1, n + 1}
(b) {n}
(c) {n, n + 1}
(d) {n – 1, n}
47. If the number of elements in (A ~ B) ~ C, (B ~ C)
~ A, (C ~ A) ~ B and A « B « C is 10, 15, 20,
and 5 respectively then the number of elements in
(A D B) D C is
(a) 35
(b) 50
(c) 40
(d) 45
x-3
, x π - 1. Then f 2010(2014) (where
48. Let f(x) =
x +1
f n(x) = fof . . . of (x) (n times)) is
(a) 2010
(c) 4028
(b) 4020
(d) 2014
Sets, Relations and Functions 1.27
Assertion-Reason Type Questions
49. Let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
be a relation on set A = {1, 2, 3}
Statement-1: R is not an equivalence relation on A.
Statement-2: R is function from A to A.
50. Statement-1:
If A is a set with 5 elements and B is a set containing
9 elements, then the number of injective mappings
from A to B to 9! – 4!.
Statement-2: The total number of injective mappings
from a set with m elements to a set with n elements,
n!
m £ n, is
.
(n - m)!
sin x
.
51. Let f(x) = sin x + cos x, g(x) =
1 - cos x
Statement-1: f is neither an odd function nor an even
function.
Statement-2: g is an odd function.
52. Statement-1: A function f : R Æ R satisfies the equation f (x) – f (y) = x – y " x, y Œ R and f (3) = 2, then
f (xy) = xy – 1
Statement-2: f (x) = f (1/x) " x Œ R, x π 0, and
f(2) = 7/3 if f (x) =
x2 + x + 1
x2 - x + 1
.
53. Statement-1: Let A = {2, 3, 7, 9} and B = {4,
9, 49, 81} f : A Æ B is a function defined as
f (x) = x2. Then f is a bijection from A to B.
Statement-2: A function f from a set A to a set
B is a bijection if f (A) = B and f (x1) π f(x2) if
x1 π x2 for all x1, x2 Œ A and n(A) = n(B).
LEVEL 2
Straight Objective Type Questions
54. The range of the function f(x) = cos [x] for – p/2
< x < p/2 contains
(a) {–1, 1, 0}
(b) {cos 1, 1, cos 2}
(c) {cos 1, – cos1, 1}
(d) [– 1, 1].
55. The domain of the function
x-5
- 3 x + 5 is
f (x) = log10 2
x - 10 x + 24
(a) (– 5, •)
(b) (5, •)
(c) (2, 5) » (5, •)
(d) (4, 5) » (6, •)
56. If g (x) = 1 +
3
x then a function f such that
3
f (g(x)) = 3 – 3 ( x ) + x is
(a) f (x) = x3 – 3x2 + x + 5
(b) f (x) = x3 + 3x2 – x – 5
(c) f (x) = x3 – 3x2 + 5
(d) f (x) = x3 – 3x2 + 3x + 3
x
x
57. The function f (x) = x
+ + 1 is
e -1 2
(a) even
(b) periodic
(c) odd
(d) neither even nor odd
x
58. Let f and g be two functions defined by f (x) =
,
x +1
x
g (x) =
. Then (f o g)–1 (x) is equal to
1- x
(a) x
(b) 1
(c) 2x
(d) none of these
59. Let f : (–1, 1) Æ (0, p) be defined by
2x
Then
f (x) = cot -1
1 - x2
(a) f is one-one but not onto
(b) f is onto but not one-one
(c) f is both one-one and onto
(d) f is neither one-one nor onto
60. Let f : X Æ [1, 27] be a function by f (x) = 5 sin x
+ 12 cos x + 14. The set X so that f is one-one and
onto is
(a) [– p/2, p/2]
(b) [0, p]
(c) [0, p/2]
(d) none of these
1- x
. Then f o f (cos x) is equal to
61. Let f (x) =
1+ x
(a) cos 2x
(b) cos x
(c) tan 2x
(d) tan x
62. Let f (x) = sin x , then
(a) f (x) is periodic with period
2p
(b) f (x) is periodic with period p
(c) f (x) is periodic with period 4p2
(d) none of these
1.28
Complete Mathematics—JEE Main
63. Let X = Y = R ~ {1}. The function f : X Æ Y defined
x+2
by f (x) =
is
x -1
(a) one-one but not onto
(b) onto but not one-one
(c) neither one-one nor onto
(d) one-one and onto
a x + a- x
and f (x + y) + f(x – y) = K f (x)
2
f (y) then K is equal to
(a) 2
(b) 4
(c) – 2
(d) none of these
64. If f (x) =
65. If f (x) = 2 sin–1
(a) [3, 4]
(c) [– p/2, p/2]
x - 3 , then the domain of f is
(b) [– 1, 1]
(d) [6, 8]
66. Let f (x) = x |x| and g(x) =
| x | then the number of
elements in the set {x ΠR : f (x) = g (x)} is
(a) 1
(b) 2
(c) 3
(d) infinitely many
67. Which of the following functions are odd functions
Ê a x + a- x ˆ
(a) f (x) = x Á x
Ë a - a - x ˜¯
(b) f (x) =
(c) f (x) =
ax + x
ax - x
ax - 1
ax + 1
(
2
(d) f (x) = x log2 x + x + 1
)
68. Let g (x) = 1 + x – [x] and f (x) = sgn x. Then for
all x, f o g (x) is equal to
(a) x
(c) f (x)
Note
(b) 1
(d) g (x)
f(x) = sgn x = 1 if x > 0
= 0 if x = 0
= – 1 if x < 0
69. The domain of definition of the function f (x) =
log2 (log1/2 (x2 + 4x + 3)) + sin–1 (2[x]2 – 3), [x]
denotes the greatest integer £ x is
(a) 0 < x £ 1
(b) 0 £ x £ 1
(c) – • < x £ 0
(d) none of these
70. If g (f (x)) = |sin x| and f(g(x)) = (sin x )2, then f
and g may be given by
(a) f (x) = sin2 x, g(x) = x
(b) f (x) = sin x, g (x) = |x|
(c) f (x) = x2, g(x) = sin x
(d) f and g cannot be determined
71. If f (x) = 3x – 5, then f –1(x)
1
(a) is given by
3x - 5
x+5
(b) is given by
3
(c) does not exist because f is not one-one
(d) does not exist because f is not onto
72. Let f (x) = Áx – a Á, a π 0 then
(a) f (x2) = (f (x))2
(b) f ( Áx Á) = Áf (x) Á
(c) f (x + y) = f (x) + f (y)
(d) none of these
73. If n (A) = n (B) = 4 then number of bijections from
A to B is
(a) 6
(b) 24
(c) 12
(d) 18
Previous Years' AIEEE/JEE Main Questions
1. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a
relation on the set A = {1, 2, 3, 4}. The relation R is
(a) not symmetric
(b) transitive
(c) a function
(d) reflexive
[2004]
2. The range of the function is f(x) = 7 – xPx – 3 is
(a) {1, 2, 3, 4}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3}
(d) {1, 2, 3, 4, 5} [2004]
3. The domain of the function f(x) =
(a) [1, 2]
(c) [2, 3]
sin -1 ( x - 3)
(b) [2, 3]
(d) [1, 2]
4. If f : R Æ S
f(x) = sin x –
onto, then the interval of S is
(a) [0, 1]
(b) [–1, 1]
(c) [0, 3]
(d) [–1, 3]
9 - x2
is
[2004]
3 cosx + 1 is
[2004]
Sets, Relations and Functions 1.29
5. The graph of the function y = f(x) is symmetrical
about x = 2 then
(a) f (x) = f (–x)
(b) f (2 + x) = f(2 – x)
(c) f (x + 2) = f (x – 2)
(d) f (x)= –f(–x)
[2004]
6. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12),
(3, 9), (3, 12), (3, 6)} be a relation on the set
A = {3, 6, 9, 12}
The relation is
(a) an equivalence relation
(b) reflexive and symmetric only
(c) reflexive and transitive only
(d) reflexive only
[2005]
7. A real valued function f (x) satisfies the functional
equation
f (x – y) = f (x) f(y) – f(a – x) f (a + y)
where a is a given constant and f(0) = 1, f (2a – x) is
equal to
(a) f (a) + f (a – x)
(b) f (–x)
(c) –f (x)
(d) f (x)
[2005]
8. Let W denote the words in the English Dictionary.
Define the relation R by R = {(x, y) Œ W ¥ W :
the words x and y have at least one letter in common}, then R is
(a) reflexive, not symmetric and transitive.
(b) not reflexive, symmetric and transitive.
(c) reflexive, symmetric and not transition.
(d) reflexive, symmetric and transitive
[2006]
Ê p pˆ
9. The largest interval lying in Ë- , ¯ for which the
2 2
2
Êx
ˆ
function f(x) = 4–x + cos– 1 Ë - 1¯ + log cos x is
2
defined is
Ê pˆ
(a) Ë0, ¯
2
Ê p pˆ
(b) Ë- , ¯
2 2
È p pˆ
(c) Í- , ˜
Î 4 2¯
È pˆ
(d) Í0, ˜
Î 2¯
[2007]
10. Let R be the real line. Consider the following subsets of the plane R ¥ R:
S = {(x, y): y = x + 1 and 0 < x < 2}
T = {(x, y): x – y is an integer}
which of the following is true :
(a) T is an equivalence relation on R but S is not
(b) Neither S nor T is an equivalence relation.
(c) Both S and T are equivalence relation on R.
(d) S is an equivalence relation but T is not. [2008]
11. Let f : N Æ Y be a function defined as
f (x) = 4x + 3 where Y = {y ΠN : y = 4x + 3
for some x ΠN}.
Show that f is invertible and its inverse is
3y + 4
y+3
(b) g(y) = 4 +
(a) g(y) =
4
3
y+3
y-3
(c) g(y) =
(d) g(y) =
[2008]
4
4
12. If A, B and C are three sets such that A « B =
A « C and A » B = A » C, then
(a) B = C
(b) A « B = f
(c) A = B
(d) A = C
[2009]
13. Consider the following relations
R = {(x, y) | x, y are real numbers and x = wy for
some rational number w}
ÏÊ m p ˆ ¸
S = ÌÁ , ˜ ˝ {m, n, p, q are integers such that
ÓË n q ¯ ˛
n◊q π 0 and qm = pn}
Then
(a) S is an equivalence relation but R is not an
equivalence relation
(b) R and S both are equivalence relations.
(c) R is an equivalence relation but S is not an
equivalence relation.
(d) Neither R nor S is an equivalence relation.
[2010]
14. Let R be the set of real numbers.
Statement-1:
A = { (x,y) Œ R ¥ R : y – x is an integer} is
an equivalence relation on R.
Statement-2:
B = { (x, y) Œ R ¥ R : x = a y for some rational
numbers a} in an equivalence relation on R.
[2011]
15. Let f be a function defined by
f (x) = (x – 1)2 + 1, (x ≥ 1)
Statement-1:
The set {x : f (x) = f –1 (x)} = {1, 2}
Statement-2: f is a bijection and
f –1(x) = 1+
x - 1 , x ≥ 1.
[2011]
16. Consider the following relation R on the set of real
square matrices of order 3.
R = {(A,B): A = P –1 B P for some invertible
matrix P}
Statement-1: R is an equivalence relation.
Statement-2: For any two invertible 3 ¥ 3 matrices
M and N, (M N)–1 = N –1M –1
[2011]
17. Let X = {1, 2, 3, 4, 5}. The number of different
ordered pairs (Y, Z) that can be formed such that
Y Õ X, Z Õ X and Y « Z is empty is
1.30
Complete Mathematics—JEE Main
(a) 25
(c) 52
(b) 53
(d) 35.
[2012]
18. Let A and B be two sets containing 2 elements and
4 elements respectively. The number of subsets of
A ¥ B having 3 or more elements is
(a) 211
(b) 256
(c) 220
(d) 219
[2013]
19. Let P be the relation defined on the set of all real
numbers such that
P = {(a, b) : sec2a – tan2b = 1}. Then P is
(a) reflexive and symmetric but not transitive.
(b) reflexive and transitive but not symmetric.
(c) symmetric and transitive but not reflexive.
(d) an equivalence relation.
[2014]
1
3
n
È
˘
20. Let f(n) = Í +
n, where [n] denotes the greatest
Î 3 100 ˙˚
56
integer less than or equal to n. Then
to
(a) 689
(b) 1399
(c) 1287
(d) 56
 f ( n)
is equal
n =1
[2014]
21. The function f(x) = |sin 4x| + |cos 2x| is a periodic
function with period
(a) p/2
(b) 2p
(c) p
(c) p/4
[2014]
| x | -1
22. Let f : R Æ R be defined by f(x) =
then f is
| x | +1
(a) onto but not one-one
(b) both one-one and onto
(c) one-one but not onto
(d) neither one-one nor onto
[2014]
23. A relation on the set A ={x: |x|< 3, x ΠZ}, where
Z is the set of integers is defined by R = {(x, y) : y
= |x|, x π –1}. Then the number of elements in the
power set of R is
(a) 32
(b) 16
(c) 8
(d) 64
[2014]
24. Let A and B be two sets containing four and two
elements respectively. Then the number of subsets of
the set A ¥ B, each having atleast three elements is
(a) 219
(b) 256
(c) 275
(d) 510
[2015]
25. In a certain town, 25% of the families own a phone
and 15% own a car, 65% families own neither a phone
nor a car and 2000 families own both a car and a
phone. Consider the following three statements:
(a) 5% families own both a car and a phone
(b) 35% families own either a car or a phone
(c) 40,000 families live in town. Then,
(a) only (a) and (b) are correct
(b) only (a) and (c) are correct
(c) only (b) and (c) are correct
(d) all (a), (b) and (c) are correct
[2015, online]
26. Let A = {x1, x2, … x7} and B = {y1, y2, y3} be two
sets containing seven and three distinct elements respectively. Then the total number of functions f : A
Æ B that are onto, if there exist exactly three x in A
such f(x) = y2 is equal to
(b) 16.7C3
(a) 14.7C2
7
(c) 12. C2
(d) 14.7C3 [2015, online]
Ê 1ˆ
27. If f(x) + 2f Á ˜ = 3x, x π 0 and S = {x Œ R : f(x) =
Ë x¯
f(–x)}; then S
(a) is an empty set
(b) contains exactly one element
(c) contains exactly two elements
(d) contains more than two elements
[2016]
28. If the function f : [1, •[ Æ [1, •[ is defined by f(x)
= 3x(x – 1), then f–1(x) is
1
1
(a)
1 - 1 + 4 log3 x
1 + 1 + 4 log3 x
(b)
2
2
(
(c) not defined
)
(
1 x ( x -1)
2(d) ÊÁ ˆ˜
Ë 3¯
)
[2016]
1
and fn+1
1- x
(x) = f0(fn(x)), n = 0, 1, 2, … . Then the f100(3) + f1
Ê 2 ˆ + f Ê 3 ˆ is equal to
2Á
Ë 2 ˜¯
Ë 3¯
8
4
(a)
(b)
3
3
5
1
(c)
(d)
[2016, online]
3
3
29. For x Œ R, x π 0, x π 1, let f0(x) =
Previous Years' B-Architecture Entrance
Examination Questions
1. The domain of the function
f(x) =
2 x - 3 + sin x +
x - 1 is
(a) (– •, 1]
È3 ˆ
(c) Í , •˜
Î2 ¯
(b) [0, 1]
(d) [1, •]
[2006]
Sets, Relations and Functions 1.31
2. Let f : (1, •) Æ (1, •) be defined by f(x) =
Then
(a) f is 1 – 1 and onto
(b) f is 1 – 1 but not onto
(c) f is not 1 – 1 but onto
(d) f is neither 1 – 1nor onto
x+2
.
x -1
[2008]
2
2
3. Let A = {(x, y) : x > 0, y > 0, x + y = 1} and let
B = {(x, y) : x > 0, y > 0, x6 + y6 = 1} Then A « B
(a) A
(b) B
(c) f
(d) {(0, 1), (1, 0)} [2008]
4. A school awarded 38 medals in football, 15 in basketball and 20 in cricket. Suppose these medals went
to a total of 58 students and only three students got
medals in all three sports. If only 5 students got
medals in football and basketball, then the number
of medals received in exactly two of three sports is
(a) 7
(b) 9
(c) 11
(d) 13
[2008]
5. Let Q be the set of all rational numbers and R be the
relation defined as
R = {(x, y) : 1 + xy > 0, x, y ΠQ}
Then relation R is
(a) symmetric and transitive
(b) reflexive and transitive
(c) an equivalence relation
(d) reflexive and symmetric
6. The domain of the function f(x) =
(a) (– •, 30)
(c) (3, 30) » (30, •)
[2009]
1
3 - log3 ( x - 3)
is
(b) (– •, 30) » (30, •)
(d) (4, •)
[2009]
7. Let f : R Æ R be a function defined by
f(x) = x2009 + 2009x + 2009
Then f(x) is
(a) one-one but not onto
(b) not one-one but onto
(c) neither one-one nor onto
(d) one-one and onto
[2010]
È p p˘
8. Let f be a function defined on Í- , ˙ by
Î 2 2˚
4
3
f(x) = 3 cos x – 6 cos x – 6 cos2 x – 3
Then the range of f(x) is
(a) [–12, –3]
(b) [–6, –3]
(c) [–6, 3)
(d) (–12, 3]
2 –x2
9. Statement-1: The function f(x) = x e
10. Consider the following relations
R1 = {(x, y) : x and y are integers and x = ay or y =
ax for some integer a}
R2 = {(x, y): x and y are integres and ax + by = 1 for
some integers a, b}
Then
(a) R1, R2 are not equivalence relations
(b) R1, R2 are equivalence relation
(c) R1 is an equivalence relation but R2 is not
(d) R2 is an equivalence relation but R1 is not [2012]
1
11. Let f and g be functions defined by f(x) =
xŒ
x +1
R, x π – 1 and g(x) = x2 + 1, x Œ R. Then go f is
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto
[2013]
12. Let N be the set of natural number and for a ΠN,
aN denotes the set {ax : x ΠN}.
If bN « cN, = dN, where b, c, d are natural numbers
greater than 1 and the greatest common divisor of b
and c is 1 then d equals
(a) max{b, c}
(b) min{b, c}
(c) bc
(d) b + c
[2014]
13. Let f(x) = (x + 1)2 – 1, x ≥ – 1, then the set
{x : f(x) = f – 1 (x)}:
(a) is an empty set
(b) contains exactly one element
(c) contains exactly two elements
(d) contains more than two elements
[2014]
14. Let f : R Æ R
e|x| - e - x
, then f is
e x + e- x
(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one (d) neither onto nor one-one
[2015]
f(x) =
15. If f is a function of real variable x satisfying f(x +
4) – f(x + 2) + f(x) = 0, then f is periodic with period
(a) 8
(b) 10
(c) 12
(d) 6
[2016]
Answers
Concept-based
[2010]
1. (d)
2. (c)
3. (b)
4. (d)
sin|x| is even
5. (b)
6. (b)
7. (d)
8. (c)
10. (c)
11. (c)
12. (c)
Statement-2: Product of two odd function is an even
function
[2011]
Level 1
9. (a)
1.32
Complete Mathematics—JEE Main
13. (d)
14. (c)
15. (a)
16. (d)
17. (b)
18. (b)
19. (a)
20. (a)
21. (d)
22. (b)
23. (d)
24. (c)
25. (a)
26. (d)
27. (b)
28. (b)
29. (b)
30. (b)
31. (b)
32. (a)
33. (c)
34. (c)
35. (d)
36. (c)
37. (b)
38. (d)
39. (b)
40. (a)
41. (d)
42. (b)
43. (c)
44. (a)
45. (b)
46. (c)
47. (b)
48. (d)
49. (c)
50. (d)
51. (b)
52. (b)
3.
4.
5.
53. (a)
Level 2
6.
54. (b)
55. (d)
56. (c)
57. (a)
58. (a)
59. (c)
60. (d)
61. (b)
62. (d)
63. (d)
64. (a)
65. (a)
66. (b)
67. (c)
68. (b)
69. (d)
70. (a)
71. (b)
72. (d)
73. (b)
Previous Years' AIEEE/JEE Main Questions
1. (a)
2. (c)
3. (b)
4. (d)
5. (b)
6. (c)
7. (b)
8. (c)
9. (d)
10. (a)
11. (d)
12. (a)
13. (a)
14. (c)
15. (a)
16. (b)
17. (d)
18. (d)
19. (c)
20. (b)
21. (a)
22. (d)
23. (b)
24. (a)
25. (d)
26. (d)
27. (c)
28. (b)
29. (c)
Previous Years' B-Architecture Entrance
Examination Questions
1.
5.
9.
13.
(c)
(d)
(b)
(c)
2.
6.
10.
14.
(a)
(c)
(c)
(d)
3.
7.
11.
15.
(c)
(d)
(d)
(c)
so R is not symmetric. If |a| £ b and |b| £ c then
|a| £ b £ |b| £ c so |a| £ c, hence a is related to
c. Thus R is transitive. If |a| £ b, and |b| £ a then
a £ |a| £ b £ |b| £ a. So a = b, R is anti symmetric.
f(g(x)) = g(f(x)) for all x Œ R ¤ f(cx + d) = g(ax
+ b)
¤ a(cx + d) + b = c(ax + b) + d ¤ ad + b = cb
+d
¤ f(d) = g(b)
For f(x) to be defined, we have |x| > x. Since |x| ≥
x " x Œ R and |x| = x for x Œ [0, •) so |x| > x if
x Œ R = (– •, 0)
The function is defined if – 1 £ 1 – 2x £ 1 ¤
– 2 £ – 2x £ 0
¤ 0 £ x £ 1.
f ( x ) + f (- x ) 1
f1(x) =
= [sinx – cosx – sinx – cosx]
2
2
= – cosx
f ( x ) - f (- x ) 1
= [sinx – cosx + sinx + cosx]
f2(x) =
2
2
= sinx
7. – 1 £ sinx £ 1 ¤ – 1 £ – sin x £ 1 ¤ 0 £ 1 – sinx
£2
-1
1+ x
Ê1 - xˆ
8. f(– x) = (– x)2 log
= x2 log Á
=
Ë 1 + x ˜¯
1- x
1- x
= – f(x).
– x2 log
1+ x
Level 1
9. ((A ~ B) » (B ~ A)) « A
= ((A ~ B) « A) » (B ~ A) « A
= (A ~ B) » f = (A ~ B) = {3, 5}
10. (A « B) « C = {2 ¥ 3 ¥ 5 x : x Œ N}
= {30x : x ΠN}
11. X Ã Y fi X « Y = X, X » Y = Y
Y Ã X fi X « Y = Y, X » Y = X.
So X « Y = X » Y fi X = Y
12. If X à Y, let y Œ Y and y œ X then either y œ A.
-/
4. (b)
8. (a)
12. (c)
Hints and Solutions
Concept-based
1. n(A » B¢) = n(A) + n(B¢) – n(A « B¢)
= n(A) + n(U) – n(B) – (n(A) – n(A « B))
= n(U) – n(B) + n(A « B)
= 12 – 6 + 2 = 8.
2. If a = – 1 than a is not related to a. So R is not
reflexive. If a = – 1, b = 2 then |a| £ b but |b| > a,
or y ΠA
So if y Œ A, then y Œ A « Y but y œ A « X
and thus A « X π A « Y.
If y œ A, then y Œ A » Y but y œ A » X
and thus A » X π A » Y
So X Õ Y, similarly Y Õ X.
fi
X = Y.
13. A = {7, 8}, B = {1, 2, 3, .... 9}
C = {1, 2, 4, 7, 8, 14, 16, ....}
B « C = {1, 2, 4, 7, 8}
A » (B « C) = {1, 2, 4, 7, 8}
14. A = {M, A, T, H, E, I, C, S}
B = {S, T, A, I, C}
B ~ A = f, So A D B = (A ~ B) » (B ~ A)
= A ~ B.
Verify (a) and (b) are not correct.
Sets, Relations and Functions 1.33
15. n(A ~ B) = n(A) – n(A « B)
fi 15 = n(A) – 16 fi n(A) = 31
n(A » B) = n(A) + n(B) – n(A « B)
fi 36 = 31 + n(B) – 16
fi n(B) = 21.
16. A » B, A ~ B, B ~ A
(A » B) ~ (A ~ B) = B
«
(A » B) ~ (B ~ A) = A
A D B = (A ~ B) » (B ~ A)
(A » B) ~ (A D B) = A « B
and (A ~ B) ~ A = f
Thus, the required no is 8.
17. n(A ~ B) = n(A) – n(A « B)
= n(B) – n(A « B) = n(B ~ A)
18. A » B = A D B » (A « B)
So A » B = A D B fi A « B = f.
19. n(P) = 60, n(M) = 58, n(P » M) = 100, N(P « M)
= n(P) + M(M) – n(P » M)
20. R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5)}
21. (1, 2) Œ R but (1, 2) œ R1 or R2 or R3 so R π R1
or R2 or R3.
22. L1 is not ^ L1, so R is not reflexive
L1 ^ L2 fi L2 ^ L1 fi R is symmetric.
L1 ^ L2 and L2 ^ L3 fi L1 || L3 fi R is not transitive.
23. See the definition of function
24. f (x) = f –1(x) fi f (f (x)) = x.
fi [(x + 1)2 – 1 + 1]2 – 1 = x.
fi (x + 1)2 = x + 1 fi x = 0 or x = –1
So S = {0, – 1}.
25. A ¥ A has n ¥ n = n2 elements and any subset of A
¥ A is a relation on A. The number of such subsets
2
is 2 n .
26. – 1 £ sin x £ 1 fi 1 £ f(x) £ 5 fi f(x) is not onto,
f (x) = f (x + 2p) fi f is not one-one and f is not
bijective.
27. In (a) a Æ 2a and a Æ – 2a, so it is not a function
In (b) for each x Œ R there is unique | x | = y Œ
R, so it is a function.
28. f(x) = x4 + 2 = y fi x = (y – 2)1/4.
fi f –1(x) = (x – 2)1/4
fi f –1 (83) = (81)1/4 = ± 3
and f –1(– 2) = (– 4)1/4 = f.
29. R is reflexive if (x, x). ΠR for all x Π{1, 2, 3}
fi (1, 1), (2, 2), (3, 3) ΠR
R is symmetric if (1, 2), (2, 3) ŒR
fi (2, 1), (3, 2) ŒR.
R is transitive if (1, 2), (2, 3) ŒR
fi (1, 3) ŒR also (3, 1) ŒR as
R is symmetric.
So the total numbers of elements is 9.
30. A « A = A π f if A π f, So R is not reflexive.
A«B=f fi B«A=f
ARB fi BRA fi R is symmetric
Note R is not transitive and not an equivalence
relation.
31. 2f(2) – 3f(1/2) = 22 = 4
2f(1/2) – 3f(2) = (1/2)2 = 1/4
fi 5(f(2) – f(1/2)) = 4 – 1/4 = 15/4
fi f(1/2) = f (2) – 3/4.
So 2f (2) – 3[f(2) – 3/4] = 4
fi
f(2) =
7
9
-4=- .
4
4
32. Applying C3 Æ C3 – (C1 + C2) we get f(x) = 0 for
all x.
33. ax–1 + ay–1 = 1
fi ay–1 = 1 – ax–1 fi (y – 1)log a = log(1–ax–1)
R.H.S is defined if 1 – ax–1 > 0
fi ax–1 < 1 = a° fi x – 1 < 0 as a > 1
fi x < 1 and the required domain is – • < x < 1.
34. If x is an integer, f(x) = 0
p
where p and q are
q
p
1
positive integers and p < q, so that f(x) =
<
p+q 2
if x is not an integer x – [x] =
1
1
and the required range is [0, ]
2
2
35. 2f(x2) + 3f(1/x2) = x2 – 1
fi 2f(1/x2) + 3f(x2) = 1/x2 – 1
Subtracting, f(x2) – f(1/x2) = 1/x2 – x2
fi f(x2) = 1/x2.
36. f(x + 3y, x – 3y) = 12xy = (x + 3y)2 – (x – 3y)2
fi f(x, y) = x2 – y2.
37. Let f(x) = y = 2x(x – 1)
fi log y = x(x – 1) log2
fi x2 – x – log2y = 0
so 0 £ f(x) <
1 ± 1 + 4 log2 y
2
fi
x=
fi
x=f
–1
(y) =
1 ± 1 + 4 log2 y
2
1 + 1 + 4 log2 y
as f –1(x) ≥ 1.
2
38. Number of one-one functions = 5 ¥ 4 ¥ 3 = 60.
39. If we take f (x)= 1, then f(y) π 1 as the function is
one-one. So f (x) π 1
If f(z) = 1, then statements 2 and 3 both are true,
so f(z) π 1 hence f (y) = 1 and f –1(1) = y.
[Note f(x) = 2, f (y) = 1, f (z) = 3)]
40. f(x + 4p) = f (x)
fi
f
–1
(x) =
sin n ( x + 4p )
sin nx
=
x + 4p
x
sin
sin
n
n
which holds if n = 2.
If
1.34
Complete Mathematics—JEE Main
41. (n, n) ΠR only for n = 1
m = n2 fi n = m2 unless m = n = 1
m = n2 and n = p2 fi m = p2 for m, n, p π 1
hence (a), (b), (c) none is true.
42. Let B, H, F be the sets of the three teams respectively
so n(B) = 21, n(H) = 26, n(F) = 29,
n(H « B) = 14 n(H « F) = 15, n(F « B) = 12,
n(B « H « F) = 8
and n(B » H » F) = n(B) + n(H) + n(F) – n(H « B)
– n(H « F) – n(F « B) + n(B « H « F) = 21 + 26
+ 29 – 14 – 15 – 12 + 8 = 43
43. An equilateral triangle cannot be obtuse angled or
right angled but is an isosceles triangle.
So E Ã I and E « I = E.
44. A = {3, 9, 27, 81, 243, 729}, B = {9, 81, 729, 6561}
A » B = {3, 9, 27, 81, 243, 729, 6561}
A « B = {9, 81, 729}
A ~ B = {3, 27, 243}
B ~ A = {6561}
A D B = (A ~ B) » (B ~ A) = {3, 27, 243, 6561}
45. For a π 0, the range of f is R. f is differentiable
function so f is one-one and only if f is monotonic.
f¢(x) = a + cos x
If a > 1, f¢(x) > 0 i.e. f is increasing
If a < –1, f¢(x) < 0 i.e. f is decreasing
Thus f is monotonic if a Œ R ~ [–1, 1].
n
46. f(x) =
 1 + [sin kx] =
n + [sin x] + [sin 2x] + ...
k =1
p
+ [sin nx]. If kx π
for any k = 1, 2 ... n then 0
2
< kx < p and k x π p/2 so 0 < sin kx < 1. Hence
p
[sin kx] = 0, k = 1, 2 ... n. i.e. f(x) = n. If kx =
2
p
for some k then x =
, hence sin x, sin 2 x, ...,
2k
sin(k–1)x will lie between 0 and 1 so [sin j x] = 0
1 £ j £ k–1; sin k x = 1 so f(x) can be n + 1 or n.
47. (A D B) D C is disjoint union of (A ~ B) ~ C, (B ~
C) ~ A, (C ~ A) ~ B and A « B « C.
Threfore, number of elements is (A D B) D C is 10
+ 15 + 20 +5 = 50.
Ê x - 3ˆ
48. f 2(x) = f o f(x) = f( f(x)) = f Á
Ë x + 1 ˜¯
x-3
-3
x+3
= x +1
=x-3
x -1
+1
x +1
-x - 3
-3
3
x
+
Ê
ˆ
x -1
= x.
=
f 3(x) = f Á Ë x - 1 ˜¯ - x - 3
+1
x -1
f 3k(x) = x
Hence
f 2010(2014) = f 3.670 (2014) = 2014.
49. (1, 2) Œ R, But (2, 1) œ R fi R is not symmetric
and hence not an equivalence relation.
fi Statement-1 is True. Staement-2 is False as 1 Æ
1, 2, 3.
50. Statement-2 is True. Because if A has m elements,
then first elements can be mapped to n elements.
For the 2nd element the choice is n – 1 and so on.
So the total number of injective mappings is n(n –
1) (n – 2) ..... (n – m + 1) = n! / (n – m)! Which
shows that statement-1 is False.
51. f (– x) = – sin x + cos x π f(x) or – f(x), fi f is
neither odd nor even. So statement-1 is True
- sin x
g(– x) =
= – (g(x)) fi g is an odd function
1 - cos x
fi Statement-2 is also True but does not lead to
statement-1.
52. Taking y = 3, f (x) – f (3) = x – 3
fi f (x) = x – 3 + f (3) = x – 3 + 2 = x – 1
fi f (xy) = xy – 1 fi Statement –1 is True
f (x) =
x2 + x + 1
x2 - x + 1
fi f (1/x) =
1 + x + x2
1 - x + x2
= f (x)
and f (2) = 7/3 fi Statement-2 is also True, but does
not lead to statement-1.
53 . Statement-2 is true by definition of a bijective mapping, using which statement-1 is also true.
Level 2
54. – p /2 < x < p /2 fi [x] = –2, –1, 0, 1
fi f (x) = cos[x]
= cos (–2), cos (–1), cos (0), cos 1
= cos 2, cos 1, 1 cos 1
55.
x-5
> 0 and x + 5 > 0
x - 10 x + 24
fi x > 5, x > 6 fi x > 6 fi x Œ (6, •)
2
or x < 5, x > 4 fi 4 < x < 5 fi x Π(4, 5)
1/3
x =yfix =y–1
so f ( g(x)) = 3 – 3x1/3 + x
56. Let g(x) = 1 +
3
fi f ( y) = 3 – 3( y – 1) + ( y – 1)3
= y3 – 3y2 + 5
57. f (–x) =
=
-x
x
- +1
e -1 2
x
- xe x x
- +1
1 - ex 2
Sets, Relations and Functions 1.35
=
x(e x - 1 + 1) x
- +1
2
ex - 1
fix=
fi f is onto
x
x
=
+ + 1 = f ( x)
x
2
e -1
Also f (x1) = f (x2) fi x1 = x2 fi f is one-one.
64. f (x + y) + f (x – y)
Ê x ˆ
58. (fog) (x) = f (g(x)) = f Á
Ë 1 - x ˜¯
x
x
= 1- x =
=x
x
+1 x +1- x
1- x
=
a x + y + a -( x + y ) a x - y + a -( x - y )
+
2
2
=
a x (a y + a - y ) + a - x (a - y + a y )
2
=
fi (fog)–1 (x) = x
59. f (x) = cot–1
y+2
, x ΠX = R ~ {1}
y -1
2x
is clearly one-one as
1 - x2
f (x1) = f (x2)
( a x + a - x )( a y + a - y )
2
= 2. f (x) f (y) fi k = 2.
65. x – 3 ≥ 0 fi x ≥ 3 and – 1 £
fi x £ 4. fi x Œ [3, 4].
fi x1 = x2
66. x[x] =
Next, let y = f (x) fi cot y =
2x
1 - x2
fi
| x | fi x2 = x if x is an integer
x = 0, 1
x
fi y = p/2 – q = p/2 – 2tan–1x. Taking x = tan(q / 2)
so y Π(0, p) fi 0 < y < p
f (x) =
ax + x
1 - xa x
fi
f
(–
x)
=
π –f (x)
ax - x
1 + xa x
fi f is not odd
fi – p / 4 < tan–1 x < p / 4
f (x) =
fi – 1 < x < 1 fi x Œ (–1, 1)
ax - 1
1 - ax
fi
f
(–
x)
=
= –f (x) fi f is odd.
ax + 1
1 + ax
and thus f is onto.
60. As – 52 + 122 £ 5 sinx + 12 cosx £
all x
-x
67. f (x) = x a + a
fi f (–x) = f (x) fi f is not odd.
a x - a- x
2 tan(q / 2)
=
= tanq
1 - tan 2 (q / 2)
52 + 122 for
fi 14 – 13 £ 5 sin x + 2 cos x £ 14 + 13
f (x) = x log2
(x +
(
x2 + 1
)
fi f (–x)
)
= – x log - x + x 2 + 1 π f (–x).
68. g (x) = 1 + x – [x] > 0 for all x
fi 1 £ f (x) £ 27
f is not one-one as it is periodic.
Ê 1 - cos x ˆ = (tan 2 ( / ))
f
x 2
61. fof (cos x) = f Á
Ë 1 + cos x ˜¯
2
=
x -3 £ 1
1 - tan ( x / 2) = cos x.
1 + tan 2 ( x / 2)
62. Since f(x + T) = sin
T.
x + T π sin x for any value
f (x) is not periodic.
x+2
, y ΠY = R Р{1}
63. Let y = f (x) =
x -1
fi (fog) (x) = f ( g(x)) = 1.
69. The function f (x) = log2 (log1/2(x2 + 4x + 3) + sin–1[x]
is defined if
0 < x2 + 4x +3 < 1 and –1 £ x £ 2
(I)
2
x + 4x + 3 > 0 fi x > – 1 or x < –3
2
x + 4x + 3 < 1 fi –2 – 2 < x < – 2 +
(II)
2
(III)
So the domain of the function is
–1 < x < – 2 + 2 .
70. g ( f (x)) = |sin x| =
sin 2 x
which is satisfied if f (x) = sin2 x and g (x) =
x
Complete Mathematics—JEE Main
1.36
Also it satisfies f ( g (x)) = f
( x)
= (sin x )2
y+5
71. f (x) = 3x – 5 = y fi x =
=f
3
x+5
fi f – 1(x) =
3
2
2
72. f (x ) = | x – a | π ( f (x))2
–1
f (a)2 = 12 – 1 = 0 fi f (a) = 0
fi
Putting x = a, y = a, we get
f (0) = [f (a)]2 – f (0) f (2a)
1 = 0 – f(2a) fi f (2a) = –1
Next, we put x = 0, y = a to obtain
f (–a) = f (0) f (a) – f (a) f(2a) = 0
Now, putting x = 2a, y = x, we get
( y)
f (|x|) = | |x| – a| π | f (x)|
f(2a – x) = f(2a) f(x) – f(–a) f(a + x)
f (x + y) = |x + y – a| π |x – a| + |y – a|.
= (–1) f(x) – (0) f(a + x)= – f (x)
73. Required number is 4! = 24.
8. Let w ŒW, then (w, w) Œ R.
Previous Years’ AIEEE/JEE Main Questions
1. R is not symmetric as (2, 3) Œ R but (3, 2) œ R
2. We must have
7 – x ≥ 1, x – 3 ≥ 0 and 7 – x ≥ x – 3
fi x £ 6, x ≥ 3 and x £ 5
Thus,
3£x£5
\ Range of f = {4P0, 3P1, 2P2)
= {1, 3, 2}
Thus, R is not transitive.
9. 4–x
4. We have
a 2 + b 2 £ a sin x + b cos x £
2
x ΠR.
cos–1 Ê x - 1ˆ
Ë2 ¯
x
-1 £ - 1 £ 1 or for 0 £ x £ 4
2
and log cos x
3. We have –1 £ x – 3 £ 1 and 9 – x2 > 0
fi 2 £ x £ 4 and –3 < x < 3
\ domain of f is [2, 3,)
–
\ R
Also, if w1, w2, ΠW and (w1, w2) ΠR, then (w2, w1)
ΠR.
\ R is symmetric.
Next, let w1 = ink w2 = link and w3, = let, then (w1,
w2) ΠR,
(w2, w3) Œ R but (w1, w3) œ R.
cos x > 0 or for 2np –
a 2 + b2
where n ΠN.
Thus,
–
1 + 3 £ sin x –
3 cos x £
\
– 2 + 1 £ f (x) £ 2 + 1
fi
Thus, domain of f is 0 £ x <
1+ 3
– 1 £ f(x) £ 3
5. A function f(x) is symmetrical about the line x = a if
f (a – x) = f (a + x).
6. Relation R
(12, 6) œ R, R is not symmetric.
p
p
< x < 2np +
2
2
ΠR but
10. As x π x + 1, S
p
p
or È0, ˆ
ÍÎ 2 ¯
2
fi S is not an equivalence relation.
As x Рx = 0 ΠI , T
xTy fi x Рy ΠI
\
fi y Рx ΠI fi yTx
T is symmetric
Also, as (3,3) Œ R, (3, 6) Œ R fi (3, 6) ŒR;
Suppose xTy and yTz
(3, 3) ΠR, (3, 9) ΠR fi (3, 9) ΠR;
fi x Рy ΠI, y Рz ΠI
(3, 3) ΠR, (3, 12) ΠR fi (3, 12) ΠR
fi (x Рy) + (y Рz) ΠI fi x Рz ΠI fi xTz
(6, 6) ΠR, (6, 12) ΠR fi (6, 12) ΠR
\
Thus, R is transitive.
\ R is reflexive and transitive.
7. Putting x = y = 0, we get
f(0) = f (0) f (0) – f (a) f(a)
T is transitive
11. As f is one-one and onto, f is invertible. So
1
y = 4x + 3 fi x = (y – 3)
4
1
Thus, inverse of f is g(y) = (y – 3).
4
Sets, Relations and Functions 1.37
12. B = ((A » B) – A) » (A « B)
= ((A » C) – A) » (A « C) = C.
13. Note that R is not symmetric as (0, 1) ΠR but (1, 0)
œ R.
m p
Next, note that qm = pm ¤
=
n q
m p
m p
=
Thus, ÊÁ , ˆ˜ ŒS fi
n q
Ë n q¯
Clearly, S
Ê m p ˆ Œ S and Ê p r ˆ Œ S fi m = p and p = r
ÁË , ˜¯
ÁË , ˜¯
q s
n q
q s
n q
m r
m r
fi Ê , ˆ ŒS
=
Ë
n s
n s¯
Thus S is transitive so is an equivalence relation.
14. A is
fi there exist non-singular matrices P and Q such that
A = P–1BP and B = Q–1CQ
\
A = P–1 (Q–1 CQ)P
= (P–1Q–1) C(QP)
= (QP)–1 C(QP)
[using statement-2]
fi (A, C) ΠR
Let x ΠR. (x, x) ΠA as x Рx = 0 is an integer.
A is symmetric
Suppose (x, y) ΠA fi x Рy is an integer
fi y – x is an integer
fi (y, x) ΠA
A is transitive
Let (x, y) ΠA and (y, z) ΠA
fi x – y and y – z are integers
fi (x – y) + (y – z) is an integer
fi x – z is an integer.
fi (x, z) ΠA
Thus, A is an equivalence relation.
B is not symmetric
Note that (0, 3 ) Œ B as 0 = (0) 3 and 0 is a rational number, but ( 3 , 0) œ B as there is no rational
number a such that 3 = (a) (0).
\ Statement-1 is true but statement-2 is false.
TIP As statement-2 is false, only possible answer is
(d).
15. Let y = f(x) = (x – 1)2 + 1
fi y – 1 = (x – 1)2
Thus, f
–1
fi x = 1 + y -1 , y ≥ 1.
(x) = 1 + x -1 , x ≥ 1.
Now, f(x) = f
–1
(x)
fi (x – 1)2 + 1 =
fi
16. As A = I AI, we get
(A, A) ΠR for all A.
\ R
Next, assume (A, B) ΠR
fi there exists a non-singular matrix P such that
A = P–1BP
fi B = PAP–1 = (P–1)–1AP–1
Thus, (B, A) ΠR
\ R is symmetric.
Next, assume that (A, B) ΠR and (B, C) ΠR
Next, note that
fi
–1
x -1 + 1
3/2
x -1 [(x – 1) – 1] = 0
fi x = 1, 2
\ Statement-1 is true and statement-2 is true and is
correct explanation for statement-1.
17. For each x ΠX, we have the following four choices:
(a) x ΠY and x ΠZ
(b) x œ Y and x Œ Z
(c) x Œ Y and x œZ
(d) x œ Y and x œ Z
As we do not want x Œ Y « Z, we are left with three
choices for each x ΠX.
Thus, the total number of ways in which Y and Z can
be chosen is 35.
18. A ¥ B contain (2) (4) = 8 elements.
The number of subsets of A ¥ B having 3 or more
elements
= Number of subsets of A ¥ B – [number of subsets
with at most 2 elements]
= 28 – (8C1 + 8C1 + 8C2) = 219
19. P = {(a, b) : sec2a – tan2b = 1}
P
p/2, p/2) œP.
P is symmetric.
Suppose (a, b) ΠP, then
sec2a – tan2b = 1
fi
(1 + tan2a) – (sec2b – 1) = 1
fi
sec2a – tan2b = 1
fi
(b, a) ΠP.
Suppose (a, b) ΠP, (b, c) ΠP
fi sec2a – tan2b = 1 and sec2b – tan2c = 1
Adding, we get
fi sec2a + sec2b – tan2b – tan2c = 2
1.38
Complete Mathematics—JEE Main
fi sec2a + 1 – tan2c = 2
fi sec2a – tan2c = 1
\ (a, c) ΠP.
20.
As 0.05x = 2000 fi x = 40,000
Thus, (a), (b), (c) are all true.
1 3n
+
< 1 fi n £ 22;
3 100
1 3n
1£ +
< 2 fi 300 £ 100 + 9n < 600
3 100
fi 23 £ n £ 55;
Also,
1 3(56)
1
+
=2
3 100
75
22
56
55
Thus, Â f (n) = Â f (n) + Â f (n) + f (56)
n =1
n =1
n = 23
26. We can choose three elements out of seven in 7C3
ways. These three elements are mapped to y2. The
remaining 4 elements are to be mapped to y1 and y3.
This can be done is 24 ways. But these include two
ways in which all the four elements are mapped to
either y1 or y3. Therefore, there are 24 –2 = 14 ways
to map containing four elements to y1 and y3. Thus
the required number of onto mappings is 14.7C3.
Ê 1ˆ
27. f(x) + 2f ÁË ˜¯ = 3x
x
55
 f (n) + 2(56)
=0+
n = 23
33
=
(55 + 23) + 112 = 1399
2
21. Period |sin(4x)| is p/4 and period of |cos 2x| is p/2.
\ period of f(x) = |sin 4x| + |cos (2x)| is
Ê l cm Ê 1 , 1 ˆ ˆ p = p
ÁË
˜˜
ÁË
2 4¯ ¯
2
1
3
f ÊÁ ˆ˜ + 2 f ( x ) =
Ë x¯
x
R = {(–2, 2), (0, 0), (1, 1), (2, 2)}
As R has four elements, the power set of R contains
16 elements.
24. n(A ¥ B) = n(A)n(B) = 4.2 = 8
The number of subsets of A ¥ B which contains at
least three elements.
= 8C 3 + 8C 4 + 8C 5 + 8C 6 + 8C 7 + 8C 8
8
8
8
8
= 2 – ( C0 + C1 + C2) = 256 – (1 + 8 + 28)
= 219
25. Let total number of families in the town be x. Then
n(P) = 0.25x, n(C) = 0.15x, n(P ¢ « C ¢) = 0.65x,
n(P » C) = x – n(P¢ « C¢) = 0.35x
i.e. 35% of the families own either a car or a phone.
n(P » C) = n(P) + n(C) – n(P « C)
fi 0.35x = 0.25x + 0.15x – n(P « C)
(ii)
(i) – 2(ii) gives f(x) – 4f(x) = 3x –
6
x
2
-x
x
22. f (–1) = f(1) = 0 \ f is not one-one.
23. A = {–2, –1, 0, 1, 2}
(i)
1
, we have
x
Replacing x by
fi f(x) =
| x | -1
2
=1–
£ 1 for all x ŒR
Also f(x) =
| x|+1
| x | +1
Thus f cannot be onto.
xπ0
f(x) = f(–x) fi
2
2
4
-x = - +x fi
= 2x fi x2 = 2
x
x
x
x = ± 2 . Thus S has exactly two elements.
28. Let y = 3x(x–1), so y ≥ 1
fi log3 y = x(x – 1)
fi x2 – x – log3 y = 0
fix=
x=
1 ± 1 + 4 log3 y
2
. As x ≥ 1, so
1
È1 + 1 + 4 log3 y ˘ .
˚
2Î
1
È1 + 1 + 4 log3 y ˘ .
˚
2Î
1
1
29. f1(x) = f0(f0(x)) =
=
1 - f 0 ( x)
1
11- x
Thus f –1(x) =
=
1- x
1
= 1x
1- x -1
f2(x) = f0(f1(x)) =
1
=
1 - f1 ( x)
1
Ê 1ˆ
1 - Á1 - ˜
Ë
x¯
fi n(P « C) = 0.05x
f3(x) = f0(f2(x)) = f0(x)
i.e. 5% of the families own a car and a phone
Thus f3k(x) = f0(x), f3k+1(x) = f1(x),
=x
Sets, Relations and Functions 1.39
f3k+2(x) = f2(x) for all x Œ R ~ {0, 1}, k ≥ 0
Ê 2ˆ
Ê 3ˆ
So, f100(3) + f1 Á ˜ + f2 Á ˜
Ë 3¯
Ë 2¯
Previous Years' B-Architecture Entrance
Examination Questions
x – 3 ≥ 0, x ≥ 1
So the domain of f is È 3 , •ˆ˜ .
ÍÎ 2 ¯
fi
x1x2 + 2x2 – x1 – 2
= x1x2 + 2x1 – x2 – 2
fi 3x2 = 3x1 fi x1 = x2
So f is one-one
x+2
fi yx – y = x + 2
If y = f(x) =
x -1
fi
(y – 1)x = y + 2
y+2
Œ(1, •)
x=
y -1
fi
Thus f is onto
6
3. x + y6 = (x2 + y2)(x4 – x2y2 + y4), so if (x, y) Œ
A « B then 1 = x4 – x2y2 + y4 = (x2 + y2)2 – 3x2y2
= 1 – 3x2y2
fi x 2y 2 = 0
fi
x = 0 or y = 0 but x > 0, y > 0
so A « B = f
4. 58 = n (F » B » C )
= n(f) + n(B) + n(C) – n (F « B) – n(B « C)
– n(F « C) + n(F « B « C)
= 38 + 15 + 20 – n (F « B) – n(B « C)
– n(F « C) + 3
fi n(F « B) + n(B « C) + n(F « C) = 18
Number of medals received in exactly two of three
sports is
= n(F « B « C¢ ) + n(B « C « F¢ )
+ n(F « C « B¢)
= n(F « B) + n(B « C) + n(F « C)
– 3 n(F « B « C)
= 18 – 9 = 9
1
, z = –1
4
1
1
=
> 0 so (x, y) ΠR
2
2
1
5
=
> 0 so (y, z) ΠR
1 + yz = 1 +
4
4
1 + xz = 1 – 2 = –1 fi (x, z) œ R. Hence R is not
transitive.
1 + xy = 1 –
1
3 3 5
= 1- +1- + = .
3
2 2 3
x +2
x +2
2. If 1
= 2
x2 - 1
x1 - 1
5. Since 1 + x > 0 for all x ΠQ, so R
(x, y) ΠR fi 1 + xy > 0
fi
1 + yx > 0 fi (y, x) ΠR
Thus R is symmetric. Take x = 2, y = –
Ê 2ˆ
Ê 3ˆ
= f1(3) + f1 Á ˜ + f2 Á ˜
Ë 3¯
Ë 2¯
1. f
2
x > 3 and 3 π log3(x – 3)
6. f
fi x > 3, x – 3 π 27 i.e. x Œ (3, •) ~ {30}
= (3, 30) » (30, •}
7. f is continuous function and f¢(x) = 2009 x2008 + 2009
> 0. So f is one-one and onto.
p p
8. For x Œ È - , ˘ , 0 £ cos x £ 1. So
ÍÎ 2 2 ˙˚
0 £ cos4x £ 1 and –5 £ – 2 cos2x – 2 cos x – 1 £ – 1
fi – 4 £ cos4x – 2 cos2x – 2cos x – 1 £ – 1
fi –12 £ f(x) £ – 3
2
2
9. f (–x) = (–x)2 e–(–x) sin |– x| = x2e–x sin |x| = f(x).
Product of two odd functions is an even function.
10. x = 1◊x so (x, x) Œ R1. If (x, y) Œ R1 then x = ay
or y = ax for some integer a
fi
y = ax or x = ay
fi (y, x) ΠR1
If x = ay or y = ax and z = by or y = bz
for some integer a and b then x = abz or z = abx, so
(x, z) ΠR1. Thus R1 is an equivalence relation.
Since (2, 2) œ R2 so R2
not an equivalence relation.
R2 is
2
Ê 1 ˆ + 1 1 + ( x + 1)2
1 ˆ
= Ë
=
11. gof (x) = g Ê
≥1
Ë x + 1¯
x + 1¯
( x + 1)2
so gof is not onto, gof is not one-one as gof (1) =
gof (–3).
12. Since g: cd (b, c) = 1 so bc = l.c.m (b, c). Hence
d = bc
13. y = f(x) = (x + 1)2 – 1 fi
Hence f
–1
(x) =
¤ (x + 1)2 – 1 =
x=
y +1 – 1
–1
x + 1 – 1. Now f(x) = f (x)
x + 1 – 1 fi (x + 1)4 = x + 1
fi (x + 1) [x3 + 3x2 + 3x] = 0
1.40
Complete Mathematics—JEE Main
fi (x + 1) x(x2 + 3x + 3) = 0
fi x = 0, –1
e- x - e- x
=0
e x + e- x
f cannot be one-one
x
For x ≥ 0, f(x) =
xŒR
Replace x by x + 2,
14. For x < 0, f(x) =
\
15. f(x + 4) – f(x + 2) + f(x) = 0
-x
e -e
≥0
e x + e- x
Thus f(x) cannot take negative values so range of f
cannot be R. Therefore f is not onto.
f(x + 6) – f(x + 4) + f(x + 2) = 0
Adding above two equations, we have
f(x + 6) + f(x) = 0
xŒR
Replace x by x + 6, we get
f(x + 12) + f(x + 6) = 0
fi f(x + 12) = –f(x + 6) = f(x)
f is periodic with period 12.
CHAPTER TWO
Complex Numbers
DEFINITIONS
A number of the form a + ib where a, b ΠR, the set of
real numbers, and i = -1 , is called a complex number. A
complex number can also be defined as an ordered pair of
real numbers a and b, and may be written as (a, b), where
the first number denotes the real part and the second number denotes the imaginary part. If z = a + ib, then the real
part of z is denoted by Re(z) and the imaginary part of z is
denoted by Im(z).
A complex number z is said to be purely real if Im(z) =
0 and is said to be purely imaginary if Re(z) = 0. Note that
the complex number 0 = 0 + i0 is both purely real and purely
imaginary. It is the only complex number with this property.
We denote the set of all complex numbers by C. That
is, C = {a + ib| a, b ΠR}. Two complex numbers z1 =
a1 + ib1 and z2 = a2 + ib2 are said to be equal if a1 = a2 and
b 1 = b 2.
5. Quotient: If at least one of c, d is non-zero, then quotient of a + ib and c + id is given by
(a + ib) (c - id )
a + ib
(ac + bd ) + i(bc - ad )
=
=
(c + id ) (c - id )
c + id
c2 + d 2
=
ac + bd
2
c +d
2
+i
bc - ad
c2 + d 2
CONJUGATE OF COMPLEX NUMBER
Let z = a + ib be a complex number. We define conjugate of
z, denoted by z to be the complex number a – ib. That is,
if z = a + ib, then z = a – ib.
Properties of Conjugate of a Complex Number
(i) z1 = z2 ¤ z1 = z2
(ii) ( z ) = z
(iii) z + z = 2 Re (z)
(iv) z – z = 2i Im(z)
Very Important
The main advantage of complex number is to use one
symbol z for an ordered pair (x, y) of real numbers. We
lose this advantage whenever we replace z by x + iy.
As far as possible do not write z = x + iy while solving the
problems. Putting z = x + iy should be the last resort rather
than the first option.
ALGEBRAIC OPERATIONS WITH
COMPLEX NUMBERS
1. Addition: (a + ib) + (c + id ) = (a + c) + i(b + d )
2. Subtraction: (a + ib) – (c + id ) = (a – c) + i(b – d )
3. Multiplication: (a + ib) (c + id )
= (ac – bd ) + i(ad + bc)
4. Reciprocal: If at least one of a, b is non-zero then the
reciprocal of a + ib is given by
a
b
1
a - ib
-i 2
=
= 2
2
a + ib
a + b2
a +b
(a + ib) (a - ib)
(v) z = z ¤ z is purely real
(vi) z + z = 0 ¤ z is purely imaginary
(vii) z z = [Re (z)]2 + [Im(z)]2
(viii) z1 + z2 = z1 + z2
(ix) z1 - z2 = z1 - z2
(x) z1z2 = z1 z2
z
Êz ˆ
(xi) Á 1 ˜ = 1 if z2 π 0
Ë z2 ¯
z2
(xii) If P(z) = a0 + a1 z + a2 z 2 + … + an zn
where a0, a1, … an and z are complex number, then
P ( z ) = a0 + a1 ( z ) + a2 ( z )2 + + an ( z )n
= P( z )
where P ( z ) = ao + a1z + a2 z 2 + + an z n
2.2
Complete Mathematics—JEE Main
P (z)
where P (z) and Q (z) are polyQ( z )
nomials in z, and Q(z) π 0, then
(xiii) If R(z) =
R (z) =
|az1 + bz2|2 + |bz1 – az2|2 = (a2 + b2) (|z1|2 + |z2|2)
(xii) If z1, z2 π 0, then |z1 + z2|2 = |z1|2 + |z2|2 ¤ z1/z2 is
purely imaginary.
P( z )
Q( z )
(xiii) Triangle Inequality. If z1 and z2 are two complex
numbers, then |z1 + z2| £ | z1| + |z2|.
1
Illustration
(xi) If a and b are real numbers and z1, z2 are complex
numbers, then
The equality holds if and only if z1 z2 ≥ 0.
Ê z + 3z 2 ˆ
z + 3z 2
=
ÁË z - 1 ˜¯
z -1
a1
(xiv) If z = b1
c1
a2
b2
c2
a3
a1 a2
b3 , then z = b1 b2
c3
c1 c2
In general, |z1 + z2 + ... + zn | £ | z1| + |z2| + ... + |zn| and
the equality sign holds if and only if the ratio of any
two non-zero terms is positive.
a3
b3
c3
(xiv) |z1 – z2| £ |z1| + |z2|
(xv )
z1 - z2 £ |z1| + |z2|
where ai, bi, ci (i = 1, 2, 3) are complex numbers.
(xvi) |z1 – z2| ≥ ||z1| – |z2||
MODULUS OF A COMPLEX NUMBER
(xvii) If a1, a2, a3 and a4 are four complex numbers, then
| z – a1 | + | z – a2 | + | z – a3 | + | z – a4 | ≥ max {| a1
– al | + | am – an|: l, m, n are distinct integers lying
in {2, 3, 4} and m < n}.
Let z = a + ib be a complex number. We define the modulus
or the absolute value of z to be the real number
and denote it by | z|.
Note that | z | 0 " z ΠC
a 2 + b2
Properties of Modulus
If z is a complex number, then
GEOMETRICAL REPRESENTATION OF
COMPLEX NUMBERS
A complex number z = x + iy can be represented by a point
P whose Cartesian co-ordinates are (x, y) referred to rectangular axes Ox and Oy, usually called the real and imaginary axes respectively. The plane is called the Argand
plane, complex plane or Gaussian plane. The point P (x, y)
is called the image of the complex number z and z is said to
be the affix or complex co-ordinate of point P.
(i) |z| = 0 ¤ z = 0
(ii) |z| = | z | = |– z| = |– z |
(iii) – | z| £ Re(z) £ |z|
(iv) – |z| £ Im(z) £ |z|
(v) z z = |z|2
In particular, note that if z π 0, then
Im ( z )
1
Ê 1 ˆ Re ( z )
=
and Im Ê ˆ = .
2
Ë z¯
Ë z¯
|z|
| z |2
If z1, z2 are two complex numbers, then
z
1
= 2 so that
z |z|
Re
Note
All purely real numbers lie on the real axis and all purely
imaginary numbers lie on the imaginary axis. The complex
number 0 = 0 + i0 lies at the origin O.
(vi) |z1 z2| = |z1| |z2|
(vii)
z1
z
= 1 , if z2 π 0
z2
z2
(viii) |z1 + z2|2 = |z1|2 + |z2|2 + z1 z2 + z1 z2
= |z1|2 + |z2|2 + 2Re( z1 z2 )
(ix) |z1 – z2|2 = |z1|2 + |z2|2 – z1z2 - z1 z2
= |z1|2 + |z2|2 – 2Re( z1 z2 )
(x) |z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2)
We have OP =
x 2 + y 2 = | z | . Thus, | z | is the length of OP.
ARGUMENT OF A COMPLEX NUMBER
If z is a non-zero complex numbers represented by point P
in the complex plane, then argument of z is the angle which
OP makes with the positive direction of the real axis. See
Fig. 2.1.
Complex Numbers 2.3
Y
Note
P (x + iy)
y
q
OP = | z |
arg (z) = q
x
X
M
O
Argument of a non-zero complex number is not unique,
since, if q is a value of the argument, then 2np + q where
n ΠI, the set of integers, are also values of the argument of
z. The value q of the argument which satisfies the inequality
– p < q £ p is called the principal value of the argument or
principal argument.
Fig. 2.1
Principal Value of the Argument for Different Positions of z in the Complex Plane
1. When z lies in the first quadrant
2. When z lies in the second quadrant
Y
z = x + iy
y
q
O
X
x
Ê yˆ
arg (z) = tan–1 Á ˜
Ë x¯
Ê yˆ
arg (z) = p - tan -1 Á ˜
Ë x¯
Fig. 2.2 (ii)
Fig. 2.2 (i)
3. When z lies in the third quadrant
Ê yˆ
arg (z) = -p + tan -1 Á ˜
Ë x¯
Fig. 2.2 (iii)
4. When z lies in the fourth quadrant
Ê yˆ
arg (z) = - tan -1 Á ˜
Ë x¯
Fig. 2.2 (iv)
Thus, if z = x + iy, then
Ïtan -1 ( y / x )
Ô -1
Ôtan ( y / x ) + p
Ô
arg ( z ) = Ìtan -1 ( y / x ) - p
Ô
Ôp / 2
ÔÓ- p /2
if x > 0
if x < 0, y ≥ 0
if x < 0, y < 0
if x = 0, y > 0
if x = 0, y < 0
Common Mistake
A usual mistake committed by the students is to take the
argument of z = x + iy as tan–1 (y/x) irrespective of the
values of x and y. Kindly remember that tan–1 (y/x) lies
in the interval (– p/2, p/2) whereas the principal value of
argument of z lies in the interval (– p, p ].
2.4
Complete Mathematics—JEE Main
POLAR FORM OF A COMPLEX NUMBER
Let z be a non-zero complex number, then we can write
z = r (cos q + i sin q )
where
r = |z| and q = arg (z).
Illustration
where r1 = |z1|, r2 = |z2|
and q1 = arg (z1), q2 = arg (z2), then
(vii) |z1 + z2|2 = |z1|2 + |z2|2 + 2|z1| |z2| cos (q1 – q2)
= r12 + r22 + 2r1r2 cos (q1 - q 2 )
|z1 + z2|2 = |z1|2 + |z2|2 ¤ cos (q1 – q2) = 0
2
¤ Oz1 is at right angles to Oz2.
(viii) |z1 – z2| = |z1|2 + |z2|2 – 2|z1| |z2| cos (q1– q2)
2
and
–1 + i =
3p
3p ˆ
Ê
2 Á cos
+ i sin ˜
Ë
4
4¯
– 1– i =
Ê 3p ˆ
Ê 3p ˆ
2 cos Á - ˜ + i sin Á - ˜
Ë 4¯
Ë 4¯
{
= r12 + r22 - 2r1r2 cos (q1 - q 2 )
}
In fact, if z = r (cos q + i sin q ), then z is also given by
z = r [cos (2kp +q ) + i sin (2kp +q)]
where k is any integer.
Euler’s Formula
The complex number cos q + i sin q is denoted by eiq or
cis q. That is
e iq = cis q = cos q + i sin q
Some Important Results Involving Argument
If z, z1 and z2 are non-zero complex numbers, then
(i) arg ( z ) = – arg (z)
(ii) *arg (z1 z2) = arg (z1) + arg (z2)
In fact,
arg (z1 z2) = arg (z1) + arg (z2) + 2kp
where
Ï0 if - p < arg ( z1) + arg ( z2 ) £ p
Ô
k = Ì1 if - 2p < arg ( z1) + arg ( z2 ) £ - p
Ô-1 if p < arg ( z ) + arg ( z ) £ 2p
Ó
1
2
(iii) *arg (z1 z2 ) = arg (z1) – arg (z2)
Êz ˆ
(iv) *arg Á 1 ˜ = arg (z1) – arg (z2)
Ë z2 ¯
In fact,
Êz ˆ
arg Á 1 ˜ = arg (z1) – arg (z2) + 2kp
Ë z2 ¯
where k is defined as in (ii) with + sign between
arg (z1) and arg (z2 ) replaced by –
(v) |z1 + z2| = |z1 – z2|
¤ arg (z1) – arg (z2) = p/2
(vi) |z1 + z2| = |z1| + |z2|
¤ arg (z1) = arg (z2)
If z1 = r1 (cos q1 + i sin q1)
and z2 = r2 (cos q2 + i sin q2)
* L.H.S. and R.H.S. may differ by a multiple of 2p.
|z1 – z2|2 = |z1|2 + |z2|2
¤ Oz1 is perpendicular to Oz2.
VECTOR REPRESENTATION OF
COMPLEX NUMBERS
We can also represent
the complex number z = x + iy by
using the vector OP joining the origin O of the complex
plane to the point P(x, y),instead
of using
point P itself.
The length of the vector OP , that is, OP is the modulus
of
z. The angle between the positive real axis and the vector
OP , more exactly, the angle through which the positive real
axismust
be rotated to cause it to have the same direction
as OP (considered positive if the rotation is counter-clockwise and negative otherwise) is the argument of the complex
number z.
GEOMETRICAL REPRESENTATION
OF ALGEBRAIC OPERATIONS ON
COMPLEX NUMBERS
Let z1 and z2 be two complex numbers represented by the
points P1(x1, y1) and P2 (x2, y2) respectively.
Sum
By definition, z1 + z2 should be represented by the point
(x1 + x2, y1 + y2). This point is nothing but the vertex P which
completes the parallelogram with the line segments joining
the origin with z1 and z2 as the adjacent sides. See Fig. 2.3
Y
P (x1 + x2, y1 + y 2)
P2 (x1, y 2)
y2
P1 (x1, y1)
M
y1
O
x2
N
L
Fig. 2.3
K
X
Complex Numbers 2.5
Note that the addition of two complex numbers z1 and z2
follows the same law of addition as that of vectors, represented both in magnitude and direction by the line segments
joining the origin and the points representing z1 and z2, for
OP1 + OP2 = OP1 + P1P = OP
Difference
We first represent – z2 by P ¢2 so that P2P ¢2 is bisected at O.
Complete the parallelogram OP1 PP ¢2. Then, it can be easily seen that P represents the difference z1 – z2. See Fig. 2.4.
As OP1 PP¢2 is a parallelogram so P1P = OP¢2. Using vector
notation, we have
z1 – z2 = OP1 – OP2 = OP1 + P2O = OP1 + OP ¢2
= OP1 + P1P = OP = P2P1
Let E be a point on the x-axis such that OE = 1 unit. (See
Fig 2.5.) Complete the triangle OP1E. Now, taking OP2 as
the base, construct a triangle OPP2 similar to OP1E so that
OP : OP1 = OP2 : OE
[ OE = 1]
i.e.,
OP = OP1 ◊ OP2
Also
– P2OP = –EOP1 = –XOP1 = q1
Thus
– XOP = q1 + q2
Hence P represents the complex number for which the modulus is r1 r2 and the argument is q1 + q2. That is, it represents
the complex number z1 z2.
Y
P2 (x2, y2)
P1 (x1, y1)
X
O
Fig. 2.5
P (x1 – x2, y1 – y 2)
P¢2 (– x2, – y2)
Fig. 2.4
Note the complex number z1 – z2 is represented by the vector P2P1, where the points P2 and P1 represent the complex
number z2 and z1 respectively.
Note that arg(z1 – z2) is the angle through which OX must
be rotated in the anticlockwise direction so that it becomes
parallel to P2P1.
Quotient
Let z1 = r1 (cos q1 + i sin q1) and z2 = r2 (cos q2 + i sin q2).
We take z2 π 0, so that r2 π 0. Now
r (cos q1 + i sin q1 )
z1
= 1
r2 (cos q 2 + i sin q 2 )
z2
=
r1
{cos (q1 – q2) + i sin (q1 –q2)}
r2
We shall use this to get a geometrical interpretation of
the quotient of a complex number by a non-zero complex
number.
Y
Product (Multiplication)
P2
= r1 r2 {cos(q1 + q2) + i sin (q1 + q2)}
Thus | z1 z2 | = r1 r2 and arg(z1 z2) = q1 + q2
This shows that modulus of the product of two complex
numbers is the product of their moduli, and the argument
of the product of any two complex numbers is the sum of
their arguments. Using this, we shall derive a geometrical
interpretation of the product of two complex numbers.
r2
Let z1 = r1(cos q1 + i sin q1) and z2 = r2 (cos q2 + i sin q2).
Then
z1 z2 = r1 r2 (cos q1 + i sin q1) (cos q2 + i sin q2)
q2
O
P1
r1
q1
E
q2
X
P
Fig. 2.6
Complete Mathematics—JEE Main
2.6
Y
Let P1 and P2 represent z1 and z2 respectively. On OP1 construct the triangle OPP1 similar to OEP2, where E lies on
the x-axis and OE = 1 unit. (See Fig. 2.6.)
Now,
Also
fi
OP : OE = r1 : r2
OP =
z4
r1
r2
z2
q = arg
– XOP = q1 – q2
z1
The point P thus represents the quotient z1/z2, since its modulus is r1/r2 and its arguments is q1 – q2.
Note that if q1 and q2 are the principal values of arg z1 and
arg z2 then q1 + q2 is not necessarily the principal value of
arg(z1 z2), nor is q1 – q2 necessarily the principal value of
arg(z1/z2).
Êz -z ˆ
Interpretation of arg Á 3 1 ˜
Ë z2 - z1 ¯
If z1, z2, z3 are the vertices of a triangle ABC described in the
counter-clockwise sense, then
– z2
3 – z4
1
I
JK
z3
O
Remark
Fz
GH z
X
Fig. 2.8
Corollary The line joining z4 and z3 is inclined at 90° to the
line joining z2 and z1 if
Ê z1 - z2 ˆ
p
=±
arg Á
˜
Ë z3 - z4 ¯
2
i.e., if z1 – z2 = ± ik(z3 – z4), where k is a non-zero real number. (Fig. 2.9).
Êz -z ˆ
(i) arg Á 3 1 ˜ = –BAC = a (say ), and
Ë z2 - z1 ¯
(ii)
z3 - z1
CA
=
(cos a + i sin a)
z2 - z1
BA
See Fig. 2.7.
C (z3)
Y
Q (z 3 – z1)
a
B (z2)
Fig. 2.9
A (z1)
a
SOME IMPORTANT GEOMETRICAL
RESULTS AND EQUATIONS
P (z 2 – z1)
X
O
Fig. 2.7
Corollary The points z1, z2, z3 will be collinear if and only
z -z
if angle a = 0 or p, i.e., if and only if 3 1 is purely real.
z2 - z1
1. Distance Formula
Distance between A(z1) and B(z2) is given by
AB = |z2 – z1|
B (z2)
Êz -z ˆ
Interpretation of arg Á 1 2 ˜
Ë z3 - z4 ¯
Let z1, z2, z3 and z4 be four complex numbers. Then the line
joining z4 and z3 is inclined to the line joining z2 and z1 at
the following angle:
Ê z1 - z2 ˆ
arg Á
Ë z3 - z4 ˜¯
A (z1)
Fig. 2.10
2. Section Formula
The point P(z) which divides the join of the segment AB in the ratio m : n is given by
Complex Numbers 2.7
z=
mz2 + nz1
m+n
Fig. 2.11
(iii) Square
(a) the diagonals AC and BD bisect each other
¤
z1 + z3 = z2 + z4
(b) a pair of adjacent sides are equal; for instance, AD = AB
¤
|z4 – z1| = |z2 – z1|
(c) the two diagonals are equal, that is,
AC = BD ¤ |z3 – z1| = |z4 – z2|
3. Mid-point Formula
Mid-point M(z) of the segment AB is given by
1
z = (z1 + z2)
2
Fig. 2.15
Fig. 2.12
4. Condition(s) for four non-collinear A(z1), B(z2),
C(z3) and D(z4) to represent vertices of a
(i) Parallelogram
The diagonals AC and BD must bisect each other
¤
¤
1
1
(z1 + z3) = (z2 + z4)
2
2
z1 + z3 = z2 + z4
(iv) Rectangle
(a) the diagonals AC and BD bisect each other
¤
z1 + z3 = z2 + z4
(b) the diagonals AC and BD are equal
¤
|z3 – z1| = |z4 – z2|.
Fig. 2.16
Showing that four points, no three of which are collinear,
form a Parallelogram/Rhombus/Square/Rectangle
Fig. 2.13
(ii) Rhombus
(a) the diagonals AC and BD bisect each other
¤
z1 + z3 = z2 + z4, and
(b) a pair of two adjacent sides are equal, for instance, AD = AB
¤
|z4 – z1| = |z2 – z1|
Fig. 2.14
2.8
Complete Mathematics—JEE Main
5. Centroid, Incentre, Orthocentre and Circumcentre of a Triangle
Let ABC be a triangle with vertices A(z1), B(z2) and
C(z3),
(i) Centroid G (z) of the DABC is the point of concurrence of medians of DABC and is given by
1
z = (z1 + z2 + z3)
3
Also
z=
z1 (sin 2 A) + z2 (sin 2 B) + z3 (sin 2C )
sin 2 A + sin 2 B + sin 2C
Fig. 2.19
(iv) Orthocentre H(z) of the DABC is the point of concurrence of altitudes of DABC and is given by
z12
Fig. 2.17
(ii) Incentre I(z) of the DABC is the point of concurrence of internal bisectors of angles of DABC
and is given by
z=
az1 + bz2 + cz3
a+b+c
z=
z22
z32
z1 1
z1 2
z1 1
2
z2 1
z3 2
z1 1
z2 1
z3 1
z3 1
z2 1 + z2
z3 1
z1
z2
z3
or z =
(tan A) z1 + (tan B) z2 + (tan C )z3
tan A + tan B + tan C
or z =
(a sec A) z1 + (b sec B)z2 + (c sec C )z3
a sec A + b sec B + c sec C
z
z
Fig. 2.18
z
(iii) Circumcentre S(z) of the DABC is the point of
concurrence of perpendicular bisectors of sides
of DABC and is given by
z=
( z2 - z3 ) z1 2 + ( z3 - z1 ) z2 2 + ( z1 - z2 ) z3 2
z1 ( z2 - z3 ) + z2 ( z3 - z1 ) + z3 ( z1 - z2 )
z=
z1 2
z1 1
z2
2
z2 1
z3
z1
z2
z3
2
z3
z1
z2
z3
1
1
1
1
z
Fig. 2.20
Remark
In case circumcentre of DABC is at the origin, then
orthocentre of triangle is given by z1 + z2 + z3.
Important Note
It is not necessary to remember formulae for circumcentre
and orthocentre of a triangle.
Complex Numbers 2.9
Euler’s Line
where
z=
The centroid G of a triangle lies on the segment joining
the orthocentre H and the circumcentre S of the triangle. G
divides the join of H and S in the ratio 2 : 1.
1
(z1 + z2 + z3)
3
A (t 1)
p/3
p/3
B (t 2)
Fig. 2.21
Thus,
1
z G = ( z H + 2 zS )
3
6. Area of a Triangle
Area of DABC with vertices A(z1), B(z2) and C(z3) is
given by
z1
1
D = | z2
4i
z3
=
z1 1
z2 1 |
z3 1
1
Im ( z1z2 + z2 z3 + z3 z1 )
2
C (t 3)
Fig. 2.23
8. Equation of a Straight Line
(i) Non-parametric form
An equation of a straight line joining the two
points A(z1) and B(z2) is
z
z1
z2
or
or
z 1
z1 1 = 0
z2 1
z - z1
z - z1
=
z2 - z1
z2 - z1
z( z1 - z2 ) - z ( z1 - z2 ) + z1 z2 - z2 z1 = 0
Fig. 2.24
Fig. 2.22
7. Condition for Triangle to be Equilateral
Triangle ABC with vertices A(z1), B(z2) and C(z3) is
equilateral if and only if
1
1
1
+
+
=0
z2 - z3
z3 - z1 z1 - z2
¤
¤
z 21 + z22 + z 23 = z2 z3 + z3 z1 + z1 z2
z1 z2 = z2 z3 = z3 z1
z12 = z2 z3 and z22 = z1z3
¤
¤
¤
¤
1 z2
1 z3
1 z1
z3
z1 = 0
z2
z2 - z1
z3 - z2
=
z3 - z2
z1 - z3
1
1
1
+
+
=0
z - z1 z - z2 z - z3
(ii) Parametric form
An equation of the straight line joining the points
A(z1) and B(z2) is
z = tz1 + (1 – t)z2
where t is a real parameter.
(iii) General Equation of a Straight Line
The general equation of a straight line is
az + az + b = 0
where a is a non-zero complex number and b is
a real number.
9. Complex Slope of a Line
If A(z1) and B(z2) are two points in the complex
plane, then complex slope of AB is defined to be
z1 - z2
z1 - z2
Two lines with complex slopes m1 and m2 are
(i) parallel, if m1 = m2
(ii) perpendicular, if m1 + m2 = 0
m=
The complex slope of the line
az + az + b = 0 is given by -(a / a )
2.10
Complete Mathematics—JEE Main
w)
A(
10. Length of Perpendicular
from a Point to a Line
Length of perpendicular
of point A(w) from the line
az + az + b = 0
p
b
z+
az
(a ΠC Р{0}, b ΠR) is given
by
p=
z
+ i k z1
z2
=0
where k is real a parameter.
+a
(v) Equation of a circle passing through three
non-collinear points.
Let three non-collinear points be A(z1), B(z2) and
C(z3). Let P(z) be any point on the circle. Then
either
–ACB = –APB
[when angles are in the same segment]
–ACB + –APB = p
[when angles are in the opposite segment]
Fig. 2.25
aw + aw + b
2a
11. Some Results on Circle
(i) Equation of a circle
An equation of a circle with centre at z0 and
radius r is
|z – z0| = r
iq
or z = z0 + re , 0 £ q < 2p (parametric form)
or
z 1
z1 1 = 0
z2 1
zz - z0 z - z0 z + z0 z0 - r 2 = 0
Fig. 2.28
fi
Fig. 2.26
(ii) General equation of a circle
General equation of a circle is
zz + az + az + b = 0
(1)
where a is a complex number and b is a real
number such that aa - b ≥ 0
Centre of (1) is – a and its radius is
aa - b .
fi
ÈÊ z - z ˆ
arg ÍÁ 3 2 ˜
ÎË z3 - z1 ¯
or
ÈÊ z - z1 ˆ Ê z3 - z2 ˆ ˘
arg ÍÁ
˜Á
˜˙ = p
ÎË z - z2 ¯ Ë z3 - z1 ¯ ˚
(iii) Diameter form of a circle
An equation of the circle one of whose diameter is the line segment joining A(z1) and B(z2) is
(z – z1) ( z – z2 ) + ( z – z1 ) (z – z2) = 0
( z - z1 ) ( z3 - z2 )
In any case, we get
is purely
( z - z2 ) ( z3 - z1 )
real.
B (z 2)
¤
( z - z1 ) ( z3 - z2 ) ( z - z1 ) ( z3 - z2 )
=
( z - z2 ) ( z3 - z1 ) ( z - z2 ) ( z3 - z1 )
(vi) Condition for four points to be concyclic.
Four points z1, z2, z3 and z4 will lie on the same
( z4 - z1 ) ( z3 - z2 )
is purely
circle if and only if
( z4 - z2 ) ( z3 - z1 )
real.
Fig. 2.27
(iv) An equation of circle through two points
An equation of the circle passing through two
points A (z1) and B (z2) is
(z – z1) ( z – z2 ) + ( z – z1 ) (z – z2)
Ê z - z1 ˆ ˘
ÁË z - z ˜¯ ˙ = 0
2 ˚
Êz ˆ
[using arg Á 1 ˜ = arg (z1) – arg (z2) and
Ë z2 ¯
arg (z1 z2) = arg (z1) + arg (z2)]
P (z)
A (z 1)
Êz -z ˆ
Ê z - z2 ˆ
=0
arg Á 3 2 ˜ - arg Á
Ë z - z1 ˜¯
Ë z3 - z1 ¯
or
Ê z3 - z2 ˆ
Ê z - z1 ˆ
=p
+ arg Á
arg Á
˜
Ë z - z2 ˜¯
Ë z3 - z1 ¯
¤
( z1 - z3 ) ( z2 - z4 )
is purely real.
( z1 - z4 ) ( z2 - z3 )
Complex Numbers 2.11
RECOGNIZING SOME LOCI BY
INSPECTION
(i) arg (z) = a (– p < a < p)
arg (z) = a represents a ray starting at the origin
(excluding the origin) and making an angle a
with the positive direction of the real axis. See
Fig. 2.29.
For k = 1, it represents perpendicular bisector of
the segment joining A(z1) and B(z2).
(v) |z – z1| + |z – z2| = k
Let z1 and z2 be two fixed points and k be a
positive real number.
(a) If k > |z1 – z2|, then
|z –z1| + |z – z2| = k represents an ellipse with
foci at A(z1) and B(z2) and length of major
axis = k. See Fig. 2.32
C
B(z2)
A(z1)
D
CD = k
Fig. 2.32
Fig. 2.29
(ii) arg (z – z0) = a (– p < a < p)
arg (z – z0) = a represents a ray starting at the
fixed point z0 (excluding the point z0) and making an angle a with the positive direction of the
real axis.
(b) If k = |z1 – z2|, then
|z – z1| + |z – z2| = k
represents the segment joining z1 and z2.
(c) If k < |z1 – z2|, then
|z – z1 | + |z – z2 | = k
does not represent any curve in the Argand
plane.
(vi) |z – z1| – |z – z2 | = k
Let z1 and z2 be two fixed points, k be a positive
real number.
(a) If k < |z1 – z2|, then
||z – z1| – |z – z2|| = k
represents a hyperbola with foci at A(z1) and
B(z2). See Fig 2.33
Fig. 2.30
(iii) If z1 and z2 are two fixed points, then
|z – z1| = |z – z2|
represents the perpendicular bisector of the
segment joining A(z1) and B(z2).
A(z1)
Fig. 2.33
z
–
|z
|z
|
z1
(b) If k = |z1 – z2|, then ||z – z1| – |z – z2|| = k
represents the straight line joining A(z1) and
B(z2) but excluding the segment AB. See
Fig. 2.34.
–z
2|
A (z1)
B(z2)
B (z2)
A(z1)
Fig. 2.31
(iv) If z1 and z2 are two fixed points, and k > 0, k π 1
is a real number, then
z - z1
=k
z - z2
represents a circle.
B(z2)
Fig. 2.34
Remark
If k > |z1 – z2|, then
||z – z1| – |z – z2|| = k
does not represent any curve in the Argand plane.
2.12
Complete Mathematics—JEE Main
(vii) If z1 and z2 are two fixed points, then
2
2
|z – z1| + |z – z2| = |z1– z2|
2
represent a circle with z1 and z2 as the extremities
of a diameter. See Fig. 2.35.
z
B(z2)
(c) If a = p, then
Ê z - z1 ˆ
arg Á
= a (= p )
Ë z - z2 ˜¯
represents the straight line joining A(z1) and
B(z2) but excluding the segment AB. See
Fig. 2.38
A(z1)
B(z2)
Fig. 2.38
A(z1)
(d) If a = 0, then
Fig. 2.35
Ê z - z1 ˆ
arg Á
= a ( = 0)
Ë z - z2 ˜¯
Ê z - z1 ˆ
=a
(viii) arg Á
Ë z - z2 ˜¯
Let z1 and z2 be two fixed points, and a be a real
number such 0 £ a £ p.
(a) If 0 < a < p and a π p /2, then
represents the segment joining A(z1) and B(z2)
see Fig. 2.39.
A(z1)
Fig. 2.39
Ê z - z1 ˆ
=a
arg Á
Ë z - z2 ˜¯
represents a segment of the circle passing through
A(z1) and B(z2). See Fig. 2.36.
B(z2)
GEOMETRIC INTERPRETATION OF
MULTIPLYING A COMPLEX NUMBER
BY eia.
Let z be a non-zero complex number. We can write z in the
polar form as follows:
z = r(cos q + i sin q) = reiq
where r = |z| and arg (z) = q
We have
zeia = reiq eia = rei(q + a )
Thus, ze ia represents the complex number whose modulus is
r and argument is q + a. Geometrically, ze ia can be obtained
by rotating the segment joining O and P(z) through an angle
a in the anticlockwise direction. See Fig. 2.40
a
A(z1)
Fig. 2.36
(b) If a = p/2, then
Ê z - z1 ˆ
Ê pˆ
arg Á
= a Á= ˜
˜
Ë 2¯
Ë z - z2 ¯
Q(zeia )
represents a circle with diameter as the segment
joining A(z1) and B(z2). See Fig. 2.37.
a
O
B(z2)
P(z)
Fig. 2.37
r
q
p/2
p/2
P(z)
r
P(z)
A(z1)
B(z2)
Fig. 2.40
In particular iz is obtained by rotating OP through an angle
of p/2 in the anticlockwise direction.
Corollary If A(z1) and B(z2) are two complex number such
that angle –AOB = q, then we can write
z2 =
z2
z eiq
z1 1
Complex Numbers 2.13
z1 = r1 eia and z2 = r2 eib,
Suppose
where
(b) De Moivre’s Theorem for rational index.
If n is a rational number, then value of or one of the
values of (cosq + isinq)n is cos (nq) + i sin (nq). In
fact, if n = p/q where p, q ΠI, q > 0 and p,q have
no factors in common, then (cos q + i sin q)n has q
distinct values, one of which is cos (nq) + i sin (nq).
|z1| = r1, |z2| = r2.
z2 r2 eib r2 i ( b -a )
=
= e
z1 r1eia r1
then
Note that
b–a=q
Thus,
Note
z2 r2 iq
= e
z1 r1
fi
The values of (cos q + i sin q) p/q where p, q ΠI, q > 0,
hcf (p,q) =1 are given by
z2
z eiq
z1 1
z2 =
Èp
Èp
˘
˘
cos Í (2 kp + q )˙ + i sin Í (2 kp + q )˙
Îq
˚
Îq
˚
where k = 0, 1, 2,…, q–1.
nth Roots of Unity
By an nth root of unity we mean any complex number z
which satisfies the equation
zn = 1
(1)
Since, an equation of degree n has n roots, there are n values
of z which satisfy the equation (1). To obtain these n values
of z, we write
1 = cos (2kp) + i sin (2kp)
where k ΠI and
Ê 2 kp ˆ
Ê 2 kp ˆ
+ i sin Á
fi
z = cos Á
Ë n ˜¯
Ë n ˜¯
[using the De Moivre’s Theorem]
where
k = 0, 1, 2, …, n –1.
Fig. 2.41
Square Roots of a Complex Number
Let z = x + iy. If a + ib is a square root of z, then
(a + ib)2 = x + iy
a2 – b2 = x, 2ab = y
fi
Also, a2 + b2 =
Thus,
x 2 + y2
a= ±
b= ±
1
2
1
x 2 + y2 + x
Note
x 2 + y2 - x
We may use any n consecutive integral values to k. For
instance, in case of 3, we may take – 1, 0 and 1 and in case of
4, we may take – 1, 0, 1 and 2 or – 2, – 1, 0 and 1.
2
If y > 0, then a and b both have the same signs.
If y < 0, then a and b have opposite signs.
Notation
Illustration
3
3 + 4i = ±
1 È
Î
2
32 + 42 + 3 +
(
)
˘
32 + 42 - 3 i ˚
1
[2 2 + 2i ] = ± (2 + i)
2
Alternatively we can use De Moivre’s Theorem.
=±
DE MOIVRE’S THEOREM AND ITS
APPLICATIONS
(a) De Moivre’s Theorem for integral index.
If n is an integer, then
(cos q + i sin q)n = cos (nq) + i sin (nq)
Ê 2p ˆ
Ê 2p ˆ
Let w = cos Á ˜ + i sin Á ˜
Ë n¯
Ë n¯
By using the De Moivre’s theorem, we can write the nth
roots of unity as
1, w, w 2, …, w n –1.
Sum of the Roots of Unity is Zero
We have
1- wn
1 + w + … + wn – 1 =
1-w
n
But w = 1 as w is a nth root of unity.
\
1 + w + … + w n –1 = 0
Also, note that
nx n -1
1
1
1
+
…+
=
x -1 x -w
x - w n -1 x n - 1
2.14
Complete Mathematics—JEE Main
Writing nth Roots of Unity When n is Odd
If n = 2m + 1, then nth roots of unity are also given by
Ê 2 kp ˆ
Ê 2 kp ˆ
+ i sin Á
z = cos Á
Ë n ˜¯
Ë n ˜¯
To obtain cube roots of a real number a, we write x3 = a as
y3 = 1 where y = x/a1/3.
y3 = 1 are 1, w, w2.
Solution of
x = a1/3, a1/3 w, a1/3 w2.
\
where k = – m, – (m – 1), …, – 1, 0, 1, 2, …m.
Since
Ê 2 kp ˆ
Ê 2 kp ˆ
cos Á = cos Á
Ë n ˜¯
Ë n ˜¯
and
Ê 2 kp ˆ
Ê 2 kp ˆ
sin Á = - sin Á
Ë n ˜¯
Ë n ˜¯
we may take the roots as
Ê 2 kp ˆ
Ê 2 kp ˆ
1, cos Á
± i sin Á
Ë n ˜¯
Ë n ˜¯
where k = 1, 2, …, m.
In terms, w we may take nth roots of unity to be 1, w+1,
w+2, … w ± m.
Writing nth Roots of Unity When n is Even
If n = 2m, then nth roots of unity are given
z = + 1, + w, + w2,…+ w m –1
Ê 2p ˆ
Ê 2p ˆ
Êpˆ
Êpˆ
+ i sin Á
= cos Á ˜ + i sin Á ˜
where w = cos Á
Ë 2m ˜¯
Ë 2m ˜¯
Ë m¯
Ë m¯
Illustration
3
Ê xˆ
To obtain cube roots of –27, we write Ë ¯ = 1
-3
x = – 3, – 3w, – 3w2
fi
nth Roots of a Complex Number
Let z π 0 be a complex number. We can write z in the polar
form as follows:
z = r (cos q + i sin q)
where
r = |z| and q = arg (z). Recall – p < q £ p.
The nth root of z has n values one of which is equal to
È
Ê arg z ˆ
Ê arg z ˆ ˘
z0 = n | z | Ícos Á
˜¯ + i sin ÁË
˜ and is called as the
Ë
n
n ¯ ˙˚
Î
principal value of
we write z as
Cube Roots of Unity
n
| z | . To obtain other values of
n
|z| ,
z = r [cos (q + 2kp) + i sin (q + 2kp)]
2
Cube roots of unity are given by 1, w, w , where
-1 - 3i
Ê 2p ˆ
Ê 2p ˆ -1 + 3i
w = cos Á ˜ + i sin Á ˜ =
and w 2 =
Ë 3¯
Ë 3¯
2
2
Some Results Involving Complex Cube Root of Unity (w)
(i) w3 = 1, w = w 2 and
4
1
= w2
w
(ii) 1 + w + w2 = 0
(iii) x3 – 1 = (x – 1) (x – w) (x – w2)
fi
Ê q + 2 kp ˆ
Ê q + 2 kp ˆ ˘
1/ n È
+ i sin Á
z1 / n = r Ícos Á
˜
˜
Ë
Ë
n ¯
n ¯ ˙˚
Î
= z0 w k where k = 0, 1, 2, …, n –1.
2p
2p
+ i sin
is a complex nth root of unity.
n
n
Thus, all the nth roots of z can be obtained by multiplying the
principal value of n | z | by different roots of unity.
and w = cos
(iv) w and w2 are roots of x2 + x + 1 = 0
Rational Power of a Complex Number
(v) a3 – b3 = (a – b) (a – bw) (a – bw2)
If z is a complex number and m/n is a rational number such that m and n are relatively prime integers and
n > 0. We define
2
2
= (a – b) (aw – bw ) (aw – bw)
2
2
(vi) a + b + c2 – bc – ca – ab
= (a + bw + cw2) (a + bw2 + cw)
(vii) a3 + b3 + c3 – 3abc
= (a + b + c) (a + bw + cw2) (a + bw2 + cw)
(viii) x3+1 = (x + 1) (x + w) (x + w2)
(ix) a3 + b3 = (a + b) (a + bw) (a + bw2)
(x) Cube roots of real number a are a1/3, a1/3 w,
a1/3 w2.
z m/n =
(n z )
m
Thus, z m/n has n distinct values which are given by
mÈ Êm
ˆ
Êm
ˆ˘
z m/n = n z
ÍÎcos ÁË n (q + 2 kp )˜¯ + i sin ÁË n (q + 2 kp )˜¯ ˙˚
( )
where k = 0, 1, 2, …, n – 1.
Complex Numbers 2.15
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Solution:
101
Example 1: If a + ib =
 i k , then (a, b) equals
k =1
(a) (0, 1)
(c) (0, –1)
Ans. (a)
(b) (0, 0)
(d) (1, 1)
TIP
Note z is circum-centre of the circle passing through three
non-collinear points A (0 + i), B (0 – i) and C(–1 + 0i)
Thus, there is only one value of z.
Example 5: If z + 2 |z| = p + 4i, then Im (z) equals
(a) p
(b) 4
Solution: Write
a + ib = i + (i2 + i3 + i4 + i5) + (i6 + i7 +i8 + i9)
+ … (i98 + i99 + i100 + i101)
TIP Use
i m + i m+1 + i m+2 + i m+3 = 0 " m ŒI
to obtain
a + ib = 0 + i fi a = 0, b = 1
n
Ê1+ iˆ
Example 2: If Ë
= – 1, n ŒN, then least value of
1- i¯
n is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (b)
(c)
Ans. (b)
Ê1+ iˆ
–1 = Ë
= in
1- i¯
The least value of n is 2.
Example 6: If |z| = z + 3 – 2i, then z equals
(a) 7/6 + i
(b) – 7/6 + 2i
(c) – 5/6 + 2i
(d) 5/6 + i
Ans. (c)
Example 3: The conjugate of a complex number z is
2
. Then Re (z) equals
1- i
(a) –1
(c) 1
Ans. (c)
Solution: z =
Solution:
fi
fi
Thus,
\
(b) 0
(d) 2
2
1 - i2
=
=1+ i
1- i
1- i
Now, Re (z)= Re ( z ) = 1.
Example 4: The number of complex numbers z such
that |z – i| = |z + i| = |z + 1| is
(a) 0
(b) 1
(c) 2
(d) infinite
Ans. (b)
(d) none of these
Solution: The given equation can be written as
z = (p – 2|z|) + 4i
As p –2|z| is real, we get
Im (z) = 4
Solution: Write 1 = – i2 in the numerator, to obtain
1+ i
-i 2 + i i(1 - i )
=i
=
=
1- i
1- i
1- i
p 2 + 16
z = |z| – 3 + 2i
|z|2 = (|z| – 3)2 + 4
= |z|2 – 6|z| + 9 + 4
|z| = 13/6
z = – 5/6 + 2i
Example 7: If w (π 1) is a cube root of unity and
(1 + w2)11 = a + bw + cw2, then (a, b, c) equals
(a) (1, 1, 0)
(b) (0, 1, 1)
(c) (1, 0, 1)
(d) (1, 1, 1)
Ans. (a)
Solution:
a + bw + cw2 = (1 + w2)11 = (–w)11
= – (w3)3w2 = 1 + w
fi
[
1 + w + w2 = 0]
a = 1, b = 1, c = 0
Example 8: If x2 + y2 = 1 and x π –1 then
equals
(a) 1
(c) 2
Ans. (d)
(b) x + iy
(d) y + ix
1 + y + ix
1 + y - ix
2.16
Complete Mathematics—JEE Main
Solution:
TIP
Sometimes, it is helpful to write x2 + y2 = zz
where z = x + iy or y + ix
Put
z = y + ix, then 1 = x2 + y2 = zz
Now,
1 + y + ix
zz + z z( z + 1)
=
=
1+ z
1+ z
1 + y - ix
= z = y + ix
Example 9: If z is a non-zero complex number, then
arg (z) + arg (z) equals
(a) 0
(b) p
(c) 2p
(d) None of these
Ans. (d)
Solution: If z ΠR and z < 0, then
arg (z) = arg (z) = p
fi
arg (z) + arg (z) = 2p
Suppose z Œ C, z π 0 and z is not a negative real number.
Let arg (z) = a, where – p < a < p
In this case arg (z) = – a, so that
arg (z) + arg(z) = 0
Example 12: If w (π 1) is a complex cube root of unity
and (1 + w4)n = (1 + w8)n, then the least positive integral
value of n is
(a) 2
(b) 3
(c) 6
(d) 12
Ans. (b)
Solution:
w4 = w, w8 = w2, we get
(1 + w)n = (1 + w2)n
fi
(– w2)n = (– w)n
fi
wn = 1
\
n =3
As
Example 13: If z =
then |z| equals
(a) 2 |sin q |
(c) 1
Ans. (c)
Solution: Using
|z|2 =
Example 10: If z ΠC and 2z = |z| + i, then z equals
3 1
+ i
(a)
6 2
(c)
3 1
+ i
6 4
3 1
+ i
(b)
6 3
(d)
3 1
+ i
6 6
Ans. (a)
Solution:
|2 z|2 = |z|2 + 1
3|z|2 = 1 fi |z| = 1 / 3
fi
Thus,
z=
3 1
+ i
6 2
7
7
Ê 1 1 ˆ
Ê 1 1 ˆ
Example 11: If z = Á
+ i˜ + Á
- i˜ , then
Ë 3 2 ¯
Ë 3 2 ¯
(a) Re (z) = 0
(b) Im (z) = 0
(c) Re (z) > 0, Im (z) < 0
(d) Re (z) < 0, Im (z) > 0
Ans. (b)
Solution:
7
7
Ê 1 1 ˆ
Ê 1 1 ˆ
- i˜ + Á
+ i˜ = z
z=Á
Ë 3 2 ¯
Ë 3 2 ¯
fi z is purely real.
\
Im (z) = 0
fi
1 + cos q + i sin q
(0 < q < p /2)
sin q + i(1 + cos q )
(b) 2|cos q |
(d) |cot (q /2)|
z1
z
= 1 if z2 π 0, we get
z2
z2
(1 + cos q )2 + sin 2 q
sin 2 q + (1 + cos q )2
|z| = 1
=1
Example 14: All the roots of (z + 1)4 = z4 lie on
(a) a straight line parallel to x-axis
(b) a straight line parallel to y-axis
(c) a circle with centre at – 1 + 0i
(d) a circle with centre at 1 + i
Ans. (b)
Solution:
TIP
It is unnecessary to find roots of (z + 1)4 = z4.
If z is a root of (z + 1)4 = z4, then |(z + 1)4| = |z4|
fi
|z + 1| = |z|
fi |z – (–1)| = |z – 0|
fi z lies on the perpendicular bisector of the segment joining –1 + 0i and 0 i.e. z lies on the line Re (z) = – 1/2.
Example 15: If a (π 1) is a fifth root of unity and b (π 1)
is a fourth root of unity then
z = (1 + a) (1 + b) (1 + a2) (1 + b2) (1 + a3) (1 + b3)
equals
(a) a
(c) ab
Ans. (d)
(b) b
(d) 0
Solution: As b π 1 is a fourth root of unity,
b 4 = 1 fi (1 – b) (1 + b + b 2 + b 3) = 0
Complex Numbers 2.17
b π 1, 1 + b + b 2 (1 + b) = 0
(1 + b) (1 + b 2) = 0
z =0
Example 16: Suppose z1, z2, z3 are vertices of an equilateral triangle whose circumcentre –3 + 4i, then |z1 + z2 +
z3| is equal to
As
fi
\
(b) 10 3
(a) 5
(c) 15
Ans. (c)
(d) 15 3
Solution: As triangle is equilateral, circumcentre and
centroid of the triangle coincides, therefore,
1
( z + z + z ) = – 3 + 4i
3 1 2 3
fi
Ans. (b)
Solution: (z – 3i) (z + 3i) = 25
fi
|z – 3i|2 = 25 or |z – 3i| = 5
Now,
|z| = |(z – 3i) + 3i|
£ |z – 3i| + |3i| = 5 + 3 = 8
Example 19: If |z – 1| = |z + 1| = |z – 2i|, then value of
|z| is
(a) 1
(b) 2
(c) 5/4
(d) 3/4
Ans. (d)
Solution: z is centre of the circle passing through 1 + 0i,
y
–1 + 0i and 0 + 2i. Clearly centre lies
on the y-axis.
2i
If z = 0 + ai the centre, then
|z1 + z2 + z3| = 3 (-3)3 + 42 = 15
Example 17: If z π 0 lies on the circle |z – 1| = 1 and w
= 5/z, then w lies on
(a) a circle
(b) an ellipse
(c) a straight line
(d) a parabola
Ans. (c)
Solution: z = 5/w and |z –1| = 1
5
- 1 = 1 fi |w – 5| = |w|
w
that is, w lies on the perpendicular bisector of the segment
joining 0 and 5 + 0i.
Therefore w lies on the straight line Re (w) = 5/2
fi
Example 18: If z = 3i +
(a) 3
(c) 16
25
, then |z| cannot exceed
z + 3i
(b) 8
(d) 18
fi
fi
\
1 + a 2 = |a + 2|
1 + a2 = a2 + 4a + 4
a = 3/4
|z| = 3/4
ai
-1
1
Fig. 2.42
Example 20: The number of complex numbers
satisfying z = iz2 is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (d)
Solution: |z| = |iz2| fi |z| = |z|2
fi
|z| = 0 or |z| = 1.
If |z| = 0, then z = 0.
If |z| = 1, we get z = 1/z, so that the equation becomes
1/z = iz2
or
z3 = – i = i3 fi z = i, iw, iw2
where w (π1) is a cube root of unity.
LEVEL 1
Straight Objective Type Questions
Example 21: If z ŒC, z œ R, and a = z2 + 3z + 5, then a
cannot take value
(a) –2/5
(b) 5/2
(c) 11/4
(d) –11/5
Ans. (c)
2
Ê 3 ˆ 11
Solution: a = Ëz + ¯ +
2
4
3
11
As z œ R, z π - , thus, a π
4
2
Example 22: Suppose a, b, c ŒC, and |a| = |b| = |c| = 1
and abc = a + b + c, then bc + ca + ab is equal to
(a) 0
(b) –1
(c) 1
(d) none of these
Ans. (c)
Solution: |a|2 = |b|2 = |c|2 = 1
fi
aa = bb = cc = 1
Now,
abc = a + b + c
fi
a b c =a + b + c
2.18
fi
fi
Complete Mathematics—JEE Main
1
1 1 1
= + +
abc
a b c
bc + ca + ab = 1
Example 23: The number of complex numbers z which
satisfy z2 + 2|z|2 = 2 is
(a) 0
(b) 2
(c) 3
(d) 4
Ans. (d)
Solution: As z2 = 2(1 – |z|2) is real, z is either purely real
or purely imaginary.
If z is purely real, then
z2 = 2 (1 – z2)
fi
z=±
2
3
TIP
If z is purely imaginary, then |z|2 = –z2.
fi
fi
fi
Example 24: Suppose a ΠR and the equation z + a|z| +
2i = 0 has no solution in C, then a satisfies the relation.
(a) |a| > 1
(b) |a| ≥ 1
fi
(d) |a| ≥
2
2
Solution: z = –a |z| – 2i
|z|2 = a2 |z|2 + 4
fi
|z|2 (1 – a2) = 4
This equation has no solution if 1 – a2 £ 0
For |a| < 1, |z| =
z=
2
1 - a2
2a
1 - a2
or
|a| ≥ 1
and
–2i
Example 25: Suppose z is a complex number and n ΠN
be such that zn = (1 + z)n = 1, then the least value of n is
(a) 3
(b) 6
(c) 9
(d) 18
Ans. (b)
Solution: |z|n = |1 + z|n = 1 fi |z| = |z + 1| =1
fi |z| = 1 and z lies on the perpendicular bisector of the
segment joining 0 + 0i and –1 + 0i, that z lies on the line
Re (z) = – 1/2.
1
Let
z = - + iy , then |z| = 1
2
1
+ y2 = 1
4
fi
y= ±
3
2
Example 26: Let z π ±i be a complex number such that
1
z-i
is purely imaginary number, then z + is
z
z+i
(a) a non-zero real number other than 1
(b) a purely imaginary number
(c) a non-zero real number
(d) 0
Ans. (c)
z-i
= ik, where k ΠR.
z+i
z – i = ikz – k
z (1 – ik) = –k + i
-k + i
z=
1 - ik
Solution: Let
In this case z2 = 2(1 + z2) fi z = ± 2i
Thus, there are four complex numbers satisfying
z2 + 2|z|2 = 2
(c) |a| >
Ans. (b)
1
3
z = - ±i
= w, w2
2
2
where w is complex cube root of unity.
If z = w, wn = (1 + w)n = (–1)n w2n
fi n is even and multiple of 3. Thus, least value of n is 6.
Similarly, for z = w2.
\
Note that |z|2 =
k2 + 1
=1
1 + k2
fi
zz = 1 fi z = 1/z
1
Thus, z + = z + z, which is a real number.
z
Also,
z + z =0
fi
2Re (z) = 0 fi Re (z) = 0
fi
z = ai for some a ΠR.
But in this case
z-i
is a real number
z+i
Therefore, z + z π 0.
Example 27: The points z1, z2, z3, z4 in the complex
plane are the vertices of a parallelogram taken in order if
and only if
(a) z1 + z4 = z2 + z3
(b) z1 + z3 = z2 + z4
(c) z1 + z2 = z3 + z4
(d) None of these
Ans. (b)
Solution: See theory.
Example 28: If the complex numbers z1, z2 and z3
represent the vertices of an equilateral triangle such that
| z1 | = | z2 | = | z3 |, then
(b) z1 + z2 – z3 = 0
(a) z1 + z2 + z3 = 0
(d) z1 + z2 + z3 π 0
(c) z1 – z2 + z3 = 0
Ans. (a)
Complex Numbers 2.19
Solution: Let | z1 | = | z2 | = | z3 | = k (say),
fi z1, z2, z3 lie on a circle with centre at the origin and
radius k. As z1, z2, z3 are vertices of an equilateral triangle,
the circumcentre and the centroid of the triangle coincide.
Therefore,
1
(z1 + z2 + z3) = 0
3
fi
z1 + z2 + z3 = 0
6
Example 29: The value of S =
is
 ÊË sin
k =1
(a) – 1
(c) – i
Ans. (d)
2p k
2p k ˆ
- i cos
7
7 ¯
Solution: We have
|az1 – bz2| 2 + |bz1 + az2| 2
= a2|z1|2 + b2 |z2|2 – abz1 z2 – ab z1 z2
+ b2 |z1| 2 + a 2 |z2|2 + ba z z 2 + ba z 1 z2
= (a2 + b2 ) ( |z1| 2 + |z2|2 )
Example 32: If a and b are real numbers between 0 and
1 such that the points z1 = a + i, z2 = 1 + bi and z3 = 0 form
an equilateral triangle, then
(a) a = b = 2 – 3 should look like this
(b) 0
(d) i
(b) a = 2 – 3 , b = 3 – 1
(c) a = 3 – 1, b = 2 – 3
(d) none of these
Ans. (a)
Solution: sin Ê 2p k ˆ – i cos Ê 2p k ˆ
Ë 7 ¯
Ë 7 ¯
2p k ˆ
Ê 2p k ˆ ˘
= - i Ècos Ê
ÍÎ Ë 7 ¯ + i sin Ë 7 ¯ ˙˚
Solution: By the hypothesis 0 < a, b < 1
and
|z1 – z2| = |z2 – z3| = |z3 – z1|
fi
|(a – 1) + i (1 – b)| = |1 + ib| = |a + i|
fi
(a – 1)2 + (1 – b)2 = 1 + b2 = a2 + 1
fi
a2 – 2a + 1 – 2b = 0 and b2 = a2
As 0 < a, b < 1 and a2 = b2, we get a = b.
\
a2 – 2a + 1 – 2a = 0 fi a2 – 4a + 1 = 0
= – i w k [De Moivre’s Theorem]
where
w = cos Ê 2p ˆ + i sin Ê 2p ˆ .
Ë 7¯
Ë 7¯
7
Note that w = 1.
6
\
S = -i
 wk =
k =1
- i w (1 - w 6 )
1-w
- i (w - w 7 ) - i (w - 1)
=
=
=i
1-w
1-w
Example 30: The complex numbers sin x + i cos 2x and
cos x – i sin 2x are conjugate to each other for
1
(a) x = np, n ΠI
(b) x = Ên + ˆ p , n Œ I
Ë
2¯
(c) x = 0
(d) no value of x.
fi
fi
Example 31: If z1 and z2 are two complex numbers and
a, b are two real numbers, then | az1 – bz2 |2 + | bz1 + az2 |2
equals
(a) (a2 + b2) | z1 z2 |
(b) (a2 + b2) (z 12 + z 22 )
(c) (a2 + b2) ( | z1 |2 + | z2 |2
(d) 2ab | z1 z2 |
Ans. (c)
As 0 < a < 1, a = 2 – 3 .
Solution: As arg (z) = p/4, we can write
p
p
z = r Êcos + i sin ˆ where r = |z|
Ë
4
4¯
fi
p
p
z 2 = r 2 Êcos + i sin ˆ
Ë
2
2¯
[Using De Moivre’s Theorem]
= r (0 + i) = ir2
Re(z2) = 0
2
sin x = cos x and cos 2x = sin 2x
fi tan x = 1 and tan 2x = 1
These two equations cannot hold simultaneously.
3
Example 33: If z π 0 is a complex number such that
arg(z) = p/4, then
(b) Im(z 2 ) = 0
(a) Re(z 2) = 0
(c) Re(z 2 ) = Im(z 2 )
(d) none of these
Ans. (a)
Ans. (d)
Solution: sin x + i cos 2 x = cos x – i sin 2x
fi sin x – i cos 2x = cos x – i sin 2x
a=2±
fi
Example 34: Let z and w be two non-zero complex
numbers such that | z | = | w | and arg (z) + arg (w) = p. Then
z equal
(a) w
(b) – w
w
(c)
(d) – w
Ans. (d)
Solution: Let | z | = | w | = r and arg (w) = q, so that
arg (z) = p – q.
We have
z = r [cos(p – q) + i sin(p – q)]
2.20
Complete Mathematics—JEE Main
= r (– cosq + i sinq )
Ans. (a)
= - r (cos q + i sin q ) = - w
Solution:
z -1
Example 35: If |z| = 1 and w =
(where z π – 1),
z +1
then Re(w) equals
1
(a) 0
(b) z + 12
(c)
1
z
z + 1 z + 12
Ans. (a)
Solution: |z| = 1
2
(d)
z+
2
z – 5i
=1
z + 5i
fi
|z – 5i| = | z – (– 5i)|
fi z lies on the perpendicular bisector of the segment joining 5i and – 5i, i.e., z lies on the x-axis.
Example 38: The inequality | z – 4| < | z – 2 | represents
the region given by
(a) Re(z) ≥ 0
(b) Re(z) < 3
(c) Re(z) £ 0
(d) Re z > 3
Ans. (d)
fi z z = 1.
Y
z -1 z -1
+
2Re(w) = w + w =
z +1 z +1
=
z - 1 (1 / z ) - 1
+
z + 1 (1 / z ) + 1
O
z -1 1- z
+
=0
z +1 1+ z
Re(w) = 0
2
3
4
X
=
fi
Example 36: Let z and w be two complex numbers
such that | z | = | w | = 1 and | z + iw | = | z – i w | = 2. Then z
equals
(a) 1 or i
(b) i or – i
(c) 1 or – 1
(d) i or – 1
Ans. (c)
Solution: We have |– i w| = |– i| |w| = 1
and
|i w | = |i| w | = 1
fi – i w and i w lie on the circle |z| = 1.
As |z – (– i w)| = |z – i w | = 2 we get z and – iw, as well
as z and i w are the end points of the same diameter, with
one end point at z.
\
– iw = i w fi w + w = 0
fi w is purely imaginary.
Let
w = ik where k ΠR.
As
|w| = 1, we get |ik| = 1
fi
|k| = 1 fi k = ± 1.
\
w = ± i fi – iw = i w = ± 1
When i w = 1, then z = – 1 and
when i w = – 1 then z = 1
Example 37: The complex
numbers z = x + i y which satisfy
z - 5i
the equation
= 1, lie on
z + 5i
(a) the x – axis
(b) the straight line y = 5
(c) a circle passing
through origin
(d) none of these
y
Fig. 2.44
Solution: If z satisfies |z – 4| = |z – 2|, then z lies on the
perpendicular bisector of the segment joining z = 2 and z = 4.
i.e., |z – 4| = |z – 2| fi Re(z) = 3.
As z = 0 does not satisfy |z – 4| < |z – 2|, we get
|z – 4| < |z – 2| represents the region Re(z) > 3.
Example 39: If z1 and z2 are two complex numbers such
z -z
that 1 2 = 1, then
z1 + z2
(a) z2 = kz1, k ΠR
(c) z1 = z2
Ans. (b)
Solution: Note that at least one of z1, z2 is different
from 0. Suppose z2 π 0. We can write
z1 - z2
z /z -1
z
z
= 1 or as 1 - 1 = 1 + 1 .
= 1, as 1 2
z1 + z2
z1 / z2 + 1
z2
z2
z
– 5i
Fig. 2.43
(d) none of these
This shows that z1/z2 lies on the perpendicular bisector of
the segment joining A (– 1 + i0) and B(1 + i0) [See Theory].
Thus, z1/z2 lies on the imaginary axis.
5i
0
(b) z2 = ikz1, k ΠR
x
\
z1
= ia for some a ΠR.
z2
fi
z2 = ikz1 for some k ŒR
fi
z2 1 - i
= =
z1 ia a
Complex Numbers 2.21
Example 40: For any complex number z, the minimum
value of |z| + |z – 2i| is
(a) 0
(b) 1
(c) 2
(d) none of these
Ans. (c)
Solution: We have, for z ŒC
|2i| = |z + (2i – z)| £ |z| + |2i – z|
fi
2 £ |z| + |z – 2i|
Thus, minimum value of |z| + |z – 2i| is 2 and it is attained
when z = i.
Example 41: If x = 2 + 5i, then the value of x3 – 5x2 +
33x – 19 is equal to
(a) – 5
(b) – 7
(c) 7
(d) 10
Ans. (d)
(c)
1
tanq
2
(d) 2
Ans. (b)
Solution:
TIP
If
z=
Re(z) =
\
Re(z) =
=
Solution: x = 2 + 5i fi x – 2 = 5i
fi
(x – 2)2 = (5i)2 fi x2 – 4x + 4 = – 25
fi x2 – 4x + 29 = 0. We now divide
x3 – 5x2 + 33x – 19 by x2 – 4x + 29.
=
1
, then
w
Re (w )
w2
1 - cos q
(1 - cos q )2 + sin 2 q
1 - cos q
1 - 2 cos q + cos2 q + sin 2 q
1 - cos q
1
=
2 (1 - cos q ) 2
2z + 1
is – 4, then
iz + 1
the locus of the point representing z in the complex plane is
(a) a straight line
(b) a parabola
(c) a circle
(d) an ellipse
Ans. (c)
Example 44:
If the imaginary part of
Solution: Let z = x + iy, then
2( x + iy) + 1 (2 x + 1) + 2iy
2z + 1
=
=
i( x + iy) + 1
(1 - y) + ix
iz + 1
Thus, x3 – 5x2 + 33x – 19 = (x – 1) (x2 – 4x + 29) + 10
= (x – 1) (0) + 10 = 10
1 - iz
, then |w| = 1
Example 42: If z = x + iy and w =
z-i
implies, that, in the complex plane
(a) z lies on the imaginary axis
(b) z lies on the real axis
(c) z lies on the unit circle
(d) none of these
Ans. (b)
Solution: |w| = 1
1 - iz
=1
z-i
|(– i) (z + i)| = |z – i|
fi
fi
|– i2 – iz| = |z – i|
fi
fi
|z + i| = |z – i|
fi z lies on the perpendicular bisector of the segment joining i and – i.
fi z lies on the real axis.
Example 43: The real part of z =
(a)
1
1 - cosq
(b)
1
2
1
is
1 - cos q + i sin q
=
[(2 x + 1) + 2iy ] [(1 - y) - ix ]
(1 - y)2 + x 2
Ê 2 z + 1ˆ
As Im Á
= – 4, we get
Ë iz + 1 ˜¯
2 y(1 - y) - x(2 x + 1)
x 2 + (1 - y)2
=–4
2x2 + 2y2 + x – 2y = 4x2 + 4(y2 – 2y + 1)
fi
fi
2x2 + 2y2 – x – 6y + 4 = 0
which represents a circle.
Example 45: The area of the triangle whose vertices
are the points represented by the complex number z, iz and
z + iz is
1 2
1 2
(a)
|z|
(b)
|z|
4
8
(c)
Ans. (c)
1 2
|z|
2
(d)
1
|z|
2
2.22
Complete Mathematics—JEE Main
Solution: Area of the triangle is given by
z
1
D = | iz
4
z + iz
Ê1
3 ˆ
1
3
1 3
= 1- Á ±
i˜ = ∓
i =
+ =1
2 2
4 4
Ë2 2 ¯
z
1
-i z 1 |
z -iz 1
fi
AB = OB
Thus,
OA = OB = AB.
\ DOAB is an equilateral triangle.
Applying R3 Æ R3 – R1 – R2, we get
z
z
1
D = | iz -i z
4
0
0
=
1
z
z
1
1 | = | (-1)
|
iz -i z
4
-1
1 1 1 2
1
1
(-1) (i )z z
= z (2 ) = z 2
1 -1 4
4
2
Example 46: If w is a complex cube root of unity, then
x +1
w
w2
a root of the equation w
w2
x + w2
1
1
x +w
= 0 is
(b) x = w
(d) x = 0
(a) x = 1
(c) x = w2
Ans. (d)
Solution: Let us denote the given determinant by D.
Applying C1 Æ C1 + C2 + C3, we get
x +1+ w + w2
D = x +1+ w + w2
x +1+w +w
w
x + w2
2
1
w2
= x
x
x +w
1
w
w2
x + w2
1
1
x +w
x
Clearly D = 0 for x = 0.
Example 47: Let z1 and z2 be two non-zero complex
z
z
numbers such that 1 + 2 = 1, then the origin and points
z2 z1
represented by z1 and z2
(a) lie on a straight line
(b) form a right triangle
(c) form an equilateral triangle
(d) none of these
Ans. (c)
1
z
Solution: Let z = 1 , then z + = 1 fi z2 – z + 1 = 0
z
z2
z1 1 ± 3i
1± 3i
fi
=
z2
2
2
If z1 and z2 are represented by A and B respectively and O
be the origin, then
fi
fi
Also,
Example 48: If (1 + x + x2)n = a0 + a1x + a2x2 + º +
a2n x2n, then value of a0 + a3 + a6 + º is
(a) 1
(b) 2n
n –1
(c) 2
(d) 3n –1
Ans. (d)
Solution: Putting x = 1, w, w 2 in
(1 + x + x2) n = a 0 + a1x + a2x2 + º + a2n x 2n, we get
(1 + 1 + 1)n = a0 + a1 + a2 + a3 + º + a2n,
(1 + w + w2)n = a0 + a1w + a2w2 + a3w3 + º + a2n w2n,
and (1 + w2 + w4)n = a0 + a1w2 + a2w4 + a3w6 + º + a2n w4n
Adding the above three equations and using 1 + w +
w2 = 0, w3 = 1 we get
3n = 3(a0 + a3 + a6 + º)
fi a0 + a3 + a6 + º = 3n–1.
Example 49:
1
1 - 2 i 3 + 5i
-5
10 i , then
z = 1 + 2i
3 - 5i -10 i
11
(a)
(b)
(c)
(d)
Ans. (b)
z is purely imaginary
z is purely real
z=0
none of these
Solution: Conjugate of z equals the determinant obtained by taking conjugate of each of its element. Therefore,
1
1 + 2 i 3 - 5i
1
1 - 2 i 3 + 5i
-5
-10 i = 1 + 2 i
-5
10 i = z
z = 1- 2i
3 + 5 i 10 i
11
3 - 5 i -10 i
11
Thus, z is purely real.
Example 50:
z=
z
OA
1 ± 3i
1 3
= 1 =
=
+ =1
z2
OB
2
4 4
OA= OB
AB
z -z
z
= 2 1 = 1- 1
OB
z2
z2
Let
If (x + iy)1/3 = a + ib, then
(a) 4(a2 – b2)
x y
+ equals
a b
(b) 2(a2 – b2)
(c) 2(a2 + b2)
(d) none of these
Ans. (a)
Solution: (x + iy)1/3 = a + ib
fi
x + iy = (a + ib)3 = a3 + 3a2 (ib) + 3a(ib)2 + (ib)3
fi
x + iy = (a3 – 3ab2) + (3a2b – b3)i
Equating real and imaginary parts, we get
Complex Numbers 2.23
x = a3 – 3ab2 and y = 3a2b – b3
fi
Thus,
x
y
= a2 – 3b2 and = 3a2 – b2.
a
b
x y
+ = 4a2 – 4b2 = 4(a2 – b2)
a b
Example 51: If z e C, the minimum value of |z| + |z – i|
is attained at
(a) exactly one point
(b) exactly two points
(c) infinite number of points
(d) none of these
Ans. (c)
Solution: We have
1 = |i| = |z + (i – z)| £ |z| + |i – z|
fi
|z| + |z – i| ≥ 1
The minimum value 1 is attained at all points z = i t where
t Π[0, 1].
Example 52: For all complex numbers z1, z2 satisfying
|z1| = 12 and |z2 – 3 – 4i| = 5, the minimum value of |z1 – z2| is
(a) 0
(b) 2
(c) 7
(d) 17
Ans. (b)
Y
Solution: |z1| = 12 implies
z1
that z1 lies on the circle with
z2
centre C1 at the origin and raC2
X
C
1
dius 12 whereas |z2 – 3 – 4i| = 5
implies z2 lies on the circle with
centre at C2 (3 + 4i) and radius
5. See Fig. 2.45. The quantity
Fig. 2.45
|z1 – z2| will be least if z1 and z2
lie on the line joining C1 and C2 i.e. on the line z = (3 + 4i)t
12
(3 + 4i), we
In fact, when we take z2 = 6 + 8i and z1 =
5
12
2
obtain |z1 – z2| =
(3 + 4i ) - (6 + 8i ) = 3 + 4i = 2.
5
5
z-2
Example 53: If z lies on the circle |z – 1| = 1, then
z
equals
(a)
(b)
(c)
(d)
Ans. (d)
0
2
–1
none of these
Solution: Note that |z – 1|
= 1 represents a circle with the
Fig. 2.46
segment joining z = 0 and z = 2
+ 0i as a diameter. See Fig. 2.46. If z lies on the circle, then
z-2
is purely imaginary.
z-0
Example 54: If 1, w, º, wn–1 are the nth roots of unity,
1
1
1
+
then value of
equals
+ +
2
2 -w 2 -w
2 - w n -1
(a)
(c)
1
(b)
2n - 1
(n - 2) 2n -1
n (2 n - 1)
2n + 1
(d) none of these
2n - 1
Ans. (d)
Solution: We know that
1
1
1
1
n ( x n -1 )
+
+
=
+
+
x -1 x - w x - w2
xn - 1
x - w n -1
Putting x = 2, we get
1
1
1
n(2 n -1 )
+
=
+
+
2 - w 2 - w2
2n - 1
2 - w n -1
p
p
Example 55: If w = cos
+ i sin , then value of
n
n
1 + w + w 2 + º + wn–1 is
p
p
(b) 1 + i tan Ê ˆ
(a) 1 + i cot Ê ˆ
Ë n¯
Ë 2p ¯
(c) 1 + i
(d) none of these
Ans. (a)
Solution: We have
1 + w + w2 + º + wn–1 =
1- wn
2
=
.
1-w 1-w
Ê np ˆ
Ê np ˆ
+ i sin Á
= cos p + i sin p = – 1
as w n = cos Á
Ë n ˜¯
Ë n ˜¯
Now,
2
2(1 - w )
2(1 - w )
=
=
1 -w
(1 - w )(1 - w ) 1 - (w + w ) + w w
=
2(1 - w )
2 - 2 Re (w )
=
Im(w )
1 - Re(w ) + i Im (w )
=1+ i
1 - Re(w )
1 - Re(w )
sin
= 1+ i
()
()
p
n
p
1 - cos
n
= 1 + i cot (p/2n)
w w = |w|2 = 1]
[
=1+ i
( ) ( )
( )
p
p
cos
2n
2n
2 p
2 sin
2n
2 sin
Example 56: If | z | = 1 and z π ± 1, then all the values
z
lie on:
of
1 – z2
(a) a line not passing through the origin.
(b) | z | = 2
(c) the x – axis
(d) the y – axis
2.24
Complete Mathematics—JEE Main
Example 59: If z2 + z + 1 = 0, where z is a complex
number, then values of
Ans. (d)
Solution: As | z | = 1, we get zz = 1.
z
Let
w=
, then
1 – z2
w+ w =
=
z
1– z
2
+
2
+
z
1– z
z
1– z
=
2
z
2
z –1
z
1– z
2
+
1/ z
1 – (1 / z )2
Solution: z 2 + z + 1 = 0 fi z = w, w2.
Let
z = w then (w + w 2) 2
= (w 2 + w 4)2 = (w4 + w 8 )2
= (w5 + w10 ) 2 = 1
3
6 2
and (w + w ) = (w6 + w12)2 = 4.
Thus,
S = 12.
=0
fi
2Re(w) = 0 fi Re(w) = 0
Thus, w lies on the y-axis.
Alternate Solution
z
z
1
w=
=
=
2
2
z –z
1– z
zz – z
=
2
1 ˆ2
1 ˆ2
1ˆ 2 Ê 2 1 ˆ
Ê
Ê
+ Á z + 2 ˜ + Á z 3 + 3 ˜ +---+ Á z 6 + 6 ˜ is
Ë
Ë
Ë
z¯
z ¯
z ¯
z ¯
(a) 12
(b) 18
(c) 54
(d) 6
Ans. (a)
Ê
Ë
S= z+
1
– 2 i Im ( z )
fi w is purely imaginary, that is, w lies on the y-axis.
Example 57: The locus of the centre of a circle which
touches the circle |z – z1| = a and |z – z2| = b π a externally is
(a) an ellipse
(b) a hyperbola
(c) a circle
(d) a pair of straight lines.
Ans. (b)
Example 60: If | z + 4| £ 3, then maximum value of
| z + 1| is
(a) 4
(b) 10
(c) 6
(d) 0
Ans. (c)
Solution: | z + 4 | £ 3 represents the interior and boundary
of the circle with center at – 4 and radius equal to 3. As – 1
is an end point of a diameter of the circle, maximum possible value of | z + 1| is 6 which is attained when z = – 7.
See Fig. 2.48
Y
Solution: Suppose |z – w| = r
touches |z – z1| = a and |z – z2| = b externally.
Then |w – z1| = a + r, |w – z 2| = b + r
fi
|w – z1| – |w – z 2| = a – b
fi w lies on a hyperbola with focii at z1 and z2
–7
4
–1
O
X
a
Z1
r
W
Fig. 2.48
r
b
Z2
Fig. 2.47
Example 58: If |z2 – 1| = |z|2 + 1, then z lies on
(a) a circle
(b) the imaginary axis
(c) the real axis
(d) an ellipse
Ans. (b)
Solution:
= |z | + |– 1|
|z 2 – 1| = |z| 2 + 1 can be written as |z2 + (– 1)|
2
z2
is a non-negative real number.
–1
¤ z2 is a non-positive real number.
¤ z lies on the imaginary axis.
¤
Example 61: Let z, w be two complex numbers such
that z + i w = 0 and arg (z w) = p, then arg z equals
3p
4
p
(c)
4
Ans. (a)
p
2
5p
(d)
4
(b)
(a)
Solution: z + i w = 0
fi
Now,
arg (zw) = p
fi
fi
2arg(z) – arg(i) = p
fi
2arg(z) = p +
z – iw = 0
Ê 2ˆ
arg Á z ˜ = p
Ë i ¯
3p
p
=
2
2
fi
arg (z) =
3p
4
Complex Numbers 2.25
Example 62: If z1 + z2 + z3 = 0 and |z1| = |z2| = |z3| = 1,
2
2
2
then value of z1 + z2 + z3 equals
(a) – 1
(c) 1
Ans. (b)
(b) 0
(d) 3
2z
= – 2z1 z2 z3 ( z1 + z2 + z3 )
As z œR, z π z, therefore |z|2 = 1
= – 2z1 z2 z3(0)
=0
|z| = 1
(b) 2 + 2
3 +1
(d)
5 +1
Ans. (d)
z -
Solution:
4
4
4
£ z=2
£ |z||z|
z
z
z2 -2 z -4£0
fi
(| z | – 1)2
fi
The maximum value
z
z
-1 =
+i
z
z
£5fi|z|£1+
5.
5 + 1 is attained when z =
5 +1.
Example 67: If |w | = 2, then the set of points x +iy =
1
wlie on
w
(a) circle
(b) ellipse
(c) parabola
(d) hyperbola
Ans. (b)
is equidistant from i and – i
z
lies on the real axis.
z
fi z is real.
Thus,
Im(z) = 0.
fi
Solution:
fi
Example 64: If a, b are distinct complex numbers with
b –a
|b | = 1, then value of
equals
1–ab
(a) 1
(b) |a|
(c) 2
(d) none of these
Ans. (a)
|b | = 1 fi bb = 1
[
Example 65: Suppose z Œ C, and z œ R. If w =
|z| =
1 - z + z2
1 + z + z2
|w | = 2 fi w = 2eiq = 2(cos q + i sin q)
x +iy = w -
1
1
= 2eiq - e-iq
w
2
5
x2
y2
3
cos q, y = sin q fi
+
=1
2
9 / 4 25 / 4
2
which represents an ellipse.
fi
x=
Example 68:
cannot exceed
(a) p/2
(c) 3p/2
b –a
b –a
b (b - a )
b –a
=
=
= |b |
1 – a /b
b –a
b -a
1–ab
is a real number, then |z| equals
(c)
(1)
Solution: z = 0 clearly satisfies (1).
For z π 0, (1) can be written as
= | b | (1) = 1.
or
ŒR.
4
Example 66: If z = 2, then the maximum value of
z
| z| is equal to
(a) 1
Example 63: If z satisfies the relation
|z – i| z1| = |z + i| z1|,
then
(a) Im (z) = 0
(b) |z| = 1
(c) Re(z) = 0
(d) none of these
Ans. (a)
\
(d) 2 3
Ê1
1
1ˆ
+
= 0 – 2z1 z2 z3 Á +
Ë z1 z2 z3 ˜¯
[ |z1| = |z2| = |z3| = 1]
Solution:
3
1 + z + z2
1
1
z
fi
ŒR fi
ŒR fi z + ŒR
2
z
1 + z + 1/ z
1+ z + z
1 1 z-z
1
1
fi z-z = - =
fi z+ =z +
z z | z |2
z
z
= (z1 + z2 + z3)2 – 2 (z2 z3 + z3 z1 + z1 z2)
z
z
(c)
Ans. (a)
(b) 2
Solution: Note that z π 0 and 1 – w =
Solution: z12 + z22 + z32
fi
(a) 1
Ê 1 ˆ
If |z| = 1, z π 1, then value of arg Á
Ë 1- z ˜¯
(b) p
(d) 2p
Ans (a)
]
Solution: As |z| = 1, z π 1, z = cos q + i sin q¸– p < q,
£ p, q π 0. Now
w=
1
1
=
1- z
1- cosq - i sin q
2.26
Complete Mathematics—JEE Main
=
(1- cosq ) + i sin q
(1- cosq ) + i sin q
=
2
2
2 (1 - cosq )
(1 - cosq ) + sin q
1 i
1 i
Êqˆ
Êp qˆ
+ cot Á ˜ = + tan Á - ˜
Ë 2 2¯
Ë
¯
2 2
2 2
2
This shows that w lies on the line x = 1/2 and –p/2
< arg (w) < p/2, Arg (w) π 0, The maximum value of Arg(w)
is never attained.
(a) 13
(c) 19
Ans (b)
Solution: Let z = x + iy, then
=
z2
is real, then point represented
z -1
by the complex number z lies
(a) on circle with centre at the origin.
(b) either on the real axis or on a circle not passing
through the origin.
(c) on the imaginary axis.
(d) either on the real axis or on a circle passing through
the origin
Example 69: If z π 1,
fi
fi
\
p
p
+ isin ), a ΠR, |a| < 1,
5
5
then S = z2015 + z2016 + z2017 + … equals
Example 72: Let z = a (cos
(a)
a 2015
z -1
(b)
a 2015
1- z
(c)
z 2015
1- a
(d)
z 2015
a -1
Solution: We have |z| = |a| <1, thus
z2
Solution: As
is real, we get
z -1
S=
z2
z2
=
z -1
z -1
But
z2( z – 1) = z 2 (z – 1)
z z (z – z ) – (z – z ) (z + z ) = 0
\
¤
(z – z ) (z z – z – z ) = 0
fi
z – z = 0 or z z – z – z = 0
fi
z lies on the real axis
or z lies on a circle through the origin.
(c) ±1
3
(d) ±1
, y ΠN, and
3
Ans (b)
Solution: We have
349 |x + iy| = |3/2 +
349
3i 2 |100
x 2 + y 2 = (9/4 + 3/4)50
fi
x 2 + y 2 = 3 fi (y)
fi
1 + k 2 = 3, 3/2, 1, as y ΠN
z 2015
1- z
z2015 = a2015 [cos(403p) + i sin (403p)] = – a2015
S=
a 2015
z -1
Example 73: If z = 20i - 21 + 20i + 21 , than one of
the possible value of |arg (z)| equals
(a) p/4
(b) p/2
(c) 3p/8
(d) p
Ans (a)
100
3 i
ˆ
Example 70: If 349 (x + iy) = Ê +
Ë2 2 3 ¯
x = ky, then value of k is
(a) ±1/3
(b) ± 2 2
fi
2(4 + i) x – (3 + i) 2iy + 26i = 0
4x + y = 0, x – 3y + 13 = 0
x = –1, y = 4
|z|2 = 17
Ans (a)
Ans (b)
¤
¤
(b) 17
(d) 11
1 + k2 = 3
fi
k = ± 2 2, ± 5 2, 0
Out of the given values, we have k = ± 2 2 .
Example 71: If (4 + i) (z + z ) – (3 + i) (z – z ) + 26i =
0, then the value of |z|2 is
Solution: 20i + 21 = (5 + 2i)2
and 20i – 21 = (2 – 5i)2
\
z = ± (5 + 2i) ± (2 + 5i)
fi
z = 7(1 + i) , 3(1 – i), 3(–1 + i), 7(–1 –i)
fi
arg (z) = p/4, –p/4, 3p/4, –3p/4
Thus, one of the possible value of |arg(z)| is p/4.
Example 74: If (a + bi)11 = x + iy, where a, b, x, y ΠR,
then (b + ai)11 equals
(a) y + ix
(b) –y – ix
(c) –x – iy
(d) x + iy
Ans (b)
Solution: (a + bi )11 = x + iy
fi
(a – bi)11 = x – iy
fi
[(–i) (b + ia)]11 = – i (y + ix)
fi
(–i)11 (b + ia)11 = – i (y + ix)
As
(–i)11 = (–i)8 (–1)3 i3 = i,
we get
(b + ia)11 = – (y + ix) = – y – ix
Complex Numbers 2.27
Example 75: If a, b, x, y Œ R, w π 1, is a cube root of
unity and (a + bw)7 = x + yw, then (b + aw)7 equals
(a) y + xw
(b) – y – xw
(c) x + yw
(d) – x – yw
Solution: Taking conjugate, we get
(a + bw 2)7 = x + y w 2 fi (aw3 + bw2)7 = xw3 + yw2
w14 (aw + b)7 = w2 (xw + y)
fi
Ans (a)
(aw + b)7 = xw + y
Assertion-Reason Type Questions
Example 76: Suppose z1, z2 are two distinct complex
numbers and a, b are real numbers.
Statement-1: If z1 + z2 = a, z1 z2 = b, then arg (z1 z2) = 0
Statement-2: If z1 + z2 = a, z1 z2 = b, then z1 = z2
Ans. (c)
Solution: Statement-2 is false. This can be seen by taking z1 and z2 to be distinct real numbers.
For statement-1, take z1 = a1 + ib1, z2 = a2 + ib2. We have
b1 + b2 = 0, a1b2 + a2b1 = 0
If b2 = 0, then b1 = 0, and statement-1 is true.
If b2 π 0, then b1 = –b2 and (a1 – a2)b2 = 0
fi
a1 – a2 = 0 or a1 = a2
Thus, a2 + ib2 =
a1 – ib1 fi z2 = z1
In this case also arg (z1 z2) = arg (z1 z1)
= arg (|z1|2) = 0
Solution: Statement-2 is true. See theory
We have
z = a + (b + ic)2017 + (b – ic)2017
= a + (b – ic)2017 + (b + ic)2017
=z
fi z is real
\ Statement-1 is true and statement-2 is correct explanation for it.
Example 79: Let w π 1, be a cube root of unity, and
a, b ΠR.
Statement-1: a3 + b3 = (a + b) (aw + bw2) (aw2 + bw)
Statement-2: x3 – 1 = (x – 1) (xw2 – w) (xw – w2)
for each x ΠR.
Ans (a)
Ans (c)
Solution: We have
(xw2 – w) (xw – w2)
= x2w3 – xw2 – xw4 + w3
= x2 – (w2 + w)x + 1
= x2 + x + 1
[ w2 + w = –1]
\ (x – 1) (xw2 – w) (xw – w2)
= (x – 1) (x2 + x + 1) = x3 – 1
Thus, Statement-2 is true.
Replacing x by –x, we get
(–x)3 –1 = (–x –1) (–x w2 – w) (–xw – w2)
z -z
1 1
Solution: |z1 – z2| =
= 2 1
z1 z2
z2 z1
As z1 π z2, we get |z1 z2| = 1
fi
x3 + 1 = (x + 1) (x w2 + w) (xw + w2)
Taking conjugate of both the sides, we get
x3 + 1 = (x + 1) (x w + w2) (xw2 + w)
Example 77: Suppose z1 and z2 are two distinct nonzero complex numbers.
1 1
z1 z2
|z1| |z2| = 1
Statement-1: |z1 – z2| =
fi
1 1
z1 z2
then both of z1, z2 have modulus 1
Statement-2: |z1 – z2| =
fi
|z1| |z2| = 1
\
Statement-1 is true.
Statement-2 is false. For example take z1 = 3, z2 = 1/3.
Example 78: Suppose a, b, c ΠR.
Statement-1: If z = a + (b + ic)2017 + (b – ic)2017 then z is
real.
Statement-2: If z = z, then z is real
Ans (a)
[ w = w 2]
If b = 0, statement-1 is clearly true. Suppose b π 0.
Replacing x by a/b we get
3
Êa ˆÊa
ˆ
Ê aˆ
2ˆ Ê a 2
Ë b ¯ + 1 = Ë b + 1¯ Ë b w + w ¯ Ë b w + w¯
fi
a3 + b3 = (a + b) (aw + bw2) (aw2 + bw)
Thus, statement-1 is also true and statement-2 is a correct
explanation for it.
2.28
Complete Mathematics—JEE Main
Example 80: Let A, B, C be three set of complex
numbers as defined below:
A = {z : Im z ≥ 1}
B = {z : |z – 2 – i| = 3}
C = {z : Re ((1 –i) z)} =
2
1 b c
D1 = 1 c a
1 a b
Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get
2}
1
b
c
D1 = 0 c - b a - c
0 a-b b-c
2
Statement-1: |z + 1 – i| + |z – 5 – i| = 37
"zŒA«B«C
Statement-2: A « B « C consists of exactly one point.
= – (b – c)2 –(a – b) (a – c)
Ans (d)
= – [a2 + b2 + c2 – bc – ca – ab]
Solution: The set A consists of all points in the half
plane Im (z) ≥ 1, that is, all the points above and including
the line through S and parallel to the real axis. The set B
is the set of all points on the circle with centre at 2 + i and
radius 3, and the set C consists of all the points on the line
x + y = 2 . The regions in A, B and C intersect in exactly
one point viz. R. see Fig. 2.49.
Fig. 2.49
\ Statement-2 is true. Points P (–1 + i) and Q(5 + i) are
the end points of diameter of the circle |z – (2 + i)| = 3
Now, |z + 1 – i|2 + |z – 5 –i|2
= PR2 + QR2 = PQ2 = 36
\
Statement-1 is false.
Example 81: Let z1, z2, z3 be three distinct non-zero
complex numbers such that a = |z1|, b = |z2|, c = |z3|. and
a b c
D= b c a =0
c a b
Statement-1: If z1 + z2 + z3 = 0, then triangle with vertices
z1, z2 and z3 is an equilateral triangle.
Statement-2: Area of triangle with vertices z1, z2, and z3 is
3
z1 - z2 2
4
Ans (c)
=–
1
[(b – c)2 + (c – a)2 + (a – b)2]
2
Now,
D =0
fi (a + b + c) [(b – c)2 + (c – a)2 + (a – b)2]
As
a + b + c > 0, we get
a =b = c
fi
|z1| = |z2| = |z3|
fi z1, z2, z3 lie on a circle with centre at the origin and
radius equal to a.
If we take z1, z2 as opposite vertices of a diameter, then
triangle is a right triangle with right angle at z3, and its area
1
is |z3 – z1| |z3 – z2|
2
\ Statement-2 is false.
If z1 + z2 + z3 = 0, then |z1 – z2|2 + |z3|2
= |z1 – z2|2 + |–z1 – z2|2
= 2|z1|2 + 2|z2|2 = 4a2
fi
|z1 – z2| =
3a
Similarly, |z2 – z3| = |z3 – z1| = 3a
Thus, triangle is an equilateral triangle.
Example 82: Statement-1: If z1, z2, z3 are such that |z1| =
|z2| = |z3| = 1, then maximum value of |z2 – z3|2 + |z3 – z1|2 +
|z1 – z2|2 is 9.
Statement-2: If z1, z2, z3 are such that |z1| = |z2| = |z3| = 1,
then
Re ( z2 z3 + z3 z1 + z1 z2 ) ≥ - 3 / 2
Ans. (a)
Solution:
0 £ |z1 + z2 + z3|2
fi
0 £ |z1|2 + |z2|2 + |z3|2 + 2 Re ( z2 z3 + z3 z1 + z1 z2 )
fi
Re ( z2 z3 + z3 z1 + z1 z2 ) ≥ – 3/2
Solution: Applying C1 Æ C1 + C2 + C3,
we get
D = (a + b + c) D1 where
[
Therefore, statement-2 is true.
Next, |z2 – z3|2 + |z3 – z1|2 + |z1 – z2|2
|z1| = |z2| = |z3| = 1]
Complex Numbers 2.29
= 2(|z1|2 + |z2|2 + |z3|2) – 2 Re( z2 z3 + z3 z1 + z1z2 )
£ 2(1 + 1 + 1) + 2(3/2) = 9
Maximum value is obtained when z1 = 1, z2 = w, z3 = w 2,
where w is a cube root of unity.
Statement-1 is also true and statement-2 is correct
explanation for it.
Example 83: Statement-1: If w π 1 is a cube root of
unity and z is a complex number such that |z| = 1, then
2 + 3w + 4 z w 2
4w + 3w 2 z + 2 z
6
7
6
28
Ê 2ˆ
Ê 2ˆ
Ê 2ˆ
1 £ 2 Á ˜ + 3 Á ˜ = Á ˜ [ 2 + 2] = 6 < 1
Ë 3¯
Ë 3¯
Ë 3¯
3
A contradiction.
Thus, if z satisfies the equation z7 + 2z + 3 = 0, then 1 £ |z|
< 3/2.
fi
Thus, Statement-1 is true.
Statement-2 is false, as the given relation implies z3 – 3z2 +
1 = 0 which is satisfied by just 3 values of z where as 1 < |z|
£ 3/2 contains infinite number of points.
Example 85: Statement-1 If w π 1, is a cube root of
unity, then A2 = O, where
= 1.
Statement-2: If z1, z2 are two complex numbers, then
|z1| = |z2| ¤ z1 = z2
Ans. (c)
2 + 3w + 4 z w 2
1 2 + 3w + 4w 2 z
=
Solution:
4w + 3w 2 z + 2 z z 2 + 3w 2 + 4w z –1
Ê 1
Á
A= Á w
Áw 2
Ë
D=
w = w2, z = 1/z]
Thus, Statement-1 is true.
Statement-2 is false as |z1| = |z2| does not imply z1 =
fi |z1| = |z2|
However, z1 =
.
Example 84: Statement-1: If z is a root of the equation
x7 + 2x + 3 = 0, then 1 £ |z| < 3/2.
Statement-2: If z lies in the annular region 1 < |z| £ 3/2,
then z satisfies the
1
1
1
=1
+
+
z – 1 z – w z – w2
where w π 1 is a cube root of unity.
Ans. (c)
Solution: Suppose |z| < 1 and z7 + 2z + 3 = 0, then
3 = |– 3| = |z7 + 2z| £ |z7| + 2|z|
7
w2
1
w2ˆ
˜
1˜
w ˜¯
Statement-2 If w π 1, is a cube root of unity, then
1 2 + 3w + 4w 2 z
1
=
=
=1
z 2 + 3w + 4w 2 z | z |
[
w
fi
3 £ |z| + 2|z| < 1 + 2(1) = 3.
A contradiction.
Next, suppose that |z| ≥ 3/2 and z7 + 2z + 3 = 0, then
w = 1/z satisfies the equation 1 + 2w6 + 3w7 = 0. Now,
1 = |– 1| = |2w6 + 3w7| £ 2|w|6 + 3|w7|
x +1
w
w2
w
x + w2
1
w2
1
x +w
= x3
Ans. (b)
Solution: It is easy to show
Êy
A = Áy
Á
Ëy
2
y
y
y
yˆ
y˜ = O
˜
y¯
where
y = 1 + w + w2 = 0.
\ Statement-1 is true.
Using C1 Æ C1 + C2 + C3, we get
1
w
D = x 1 x + w2
1
1
w2
1
w
1 = x 0 x + w2 – w
x +w
0
1–w
w2
1 – w2
x + w – w2
= x[(x + w2 – w) (x + w – w2) – (1 – w) (1 – w2)]
= x[x2 – (w – w2)2 – {1 – w – w2 + 1}] = x3
\ Statement-2 is also true but is not the correct reason for
the statement-1.
2.30
Complete Mathematics—JEE Main
LEVEL 2
Straight Objective Type Questions
Example 86: Let z = cos q + isin q. Then the value of
15
 Im (z2 m - 1 ) at q = 2° is
with 0 < t < 1. If Arg (w) denotes the principal argument of
a nonzero complex number w, then which of the following
is not true?
m =1
(a)
1
sin 2∞
(b)
1
1
(c)
(d)
sin 2∞
4 sin 2∞
Ans. (d)
15
15
Solution:  Im z 2 m -1 = Im  z 2 m -1
(
)
(
m =1
z2
z
z1
(
)
2 15
( )
Fig. 2.50
) ˘˙
(a) z - z1 + z - z2 = z1 - z2
(b) Arg (z – z1) = Arg (z – z2)
Ê 1 - z 30 ˆ
˙ = Im Á
Ë z - z ˜¯
˙˚
1
(c)
˘
= Im Í
{1 - cos (30 q ) - i sin (30 q )}˙
Î -2i sin q
˚
1
1
[1 - cos (60∞)]
[1 - cos (30 q )] =
2 sin 2∞
2 sin q
1
=
4 sin 2∞
=
Example 87: Let z = x + iy be a complex number where
x and y are integers. Then the area of the rectangle whose
vertices are the roots of the equation zz 3 + zz 3 = 350 is
(a) 48
(b) 32
(c) 40
(d) 80
Ans. (a)
Solution: zz 3 + zz 3 = 350 fi zz z 2 + z 2 = 350
(
)
(
)
fi ( x 2 + y 2 ) (2 x 2 - 2 y 2 ) = 350
2
2
fi x + y ÈÎ( x - iy ) + ( x + iy ) ˘˚ =350
2
x
O
m =1
Èz 1 - z
Í
= Im Í
1 - z2
ÍÎ
È
y
1
3 sin 2∞
2
fi x4 – y4 = 175
fi x4 ≥ 175 fi x4 ≥ 256
Let us try x = ± 4
Thus, y4 =256 – 175 = 81 fi y = ± 3
\ roots of zz 3 + zz 3 = 350 are z = ± 4 ± 3i
Area of rectangle whose vertices are ± 4 ± 3i is (8) (6) = 48.
Example 88: Let z1 and z2 be two distinct complex
numbers and let z = (1 – t) z1 + tz2 for some real number t
z - z1
z - z1
z2 - z1
z2 - z1
=0
(d) Arg (z – z1) = Arg (z2 – z1)
Ans. (b)
Solution: z – z1 = t (z2 – z1)
fi
|z – z1| = t|z2 – z1| = t|z1 – z2|
(1)
[
t > 0]
z – z2 = (1 – t) (z1 – z2)
and
fi
|z – z2| = (1 – t) |z1 – z2| = (1 – t) |z2 – z1|
[
(2)
1 – t > 0]
From (1) and (2), we get
|z – z1| + |z – z2| = |z2 – z1|
Next,
z - z1
z - z1
z2 - z1
z2 - z1
=
t ( z2 - z1 )
t ( z2 - z1 )
z2 - z1
( z2 - z1 )
=0
[ R1 and R2 are identical]
Also, Arg (z – z1) = Arg (z2 – z1) since z1, z and z2 lie on the
same straight line and on the same side of z1.
Example 89: For complex numbers z1 = x1 + iy, and z2
= x2 + iy2, we write z1 « z2 if x1 £ x2 and y1 £ y2. Let z be a
complex number such that 1 « z, then
1– z
1– z
(a)
(b) 1 «
«–i
1+ z
1+ z
1– z
1+ z
(c)
(d)
«0
«0
1+ z
1– z
Ans. (c)
Complex Numbers 2.31
Solution:
Let z = x + iy. As 1 « z, we get 1 £ x and 0 £ y
1 - z (1 - x ) - iy [(1 - x ) - iy ] [(1 + x ) - iy ]
=
=
1 + z (1 + x ) + iy
(1 + x )2 + y 2
Now,
=
=
(1 - x 2 ) - y 2 - iy (1 + x + 1 - x )
n -1
(1 + x )2 + y 2
\
(1 + x )2 + y 2
1 – ( x 2 + y2 )
(1 + x )2 + y 2
=
£ 0 and
– 2y
 ÈÎ(r + 1)3 - 1˘˚
r =0
n
1 2
n (n + 1)2 - n
4
 r3 - n =
(1 + x )2 + y 2
n
£0
that |z| < 1/3, and
Example 90: The complex number z1, z2, z3 are the
vertices of an equilateral triangle. If z0 is the circumcentre of
the triangle, then z21 + z22 + z23 is equal to
(a) z02
(b) 3z02
(c) z03
(d) 3z03
r =1
(b) 1
(d) infinite
Solution: We have
n
1=
n
 ar zr £  ar
r =1
zr
r =1
n
1
 2 ÊË 3 ˆ¯
•
r
<
r =1
1
 2 ÊË 3 ˆ¯
r =1
r
=
2/3
=1
1- 2/3
A contradiction.
Solution: As the triangle with vertices z1, z2 and z3 is
an equilateral triangle, the circumcentre and the centroid of
the triangle coincides. Thus,
1
z0 = (z1 + z2 + z3)
3
(3z0)2 = z21 + z22 + z23 + 2(z2 z3 + z3 z1 + z1z2)
(1)
As the triangle is an equilateral triangle,
z 21+ z22
 ar zr = 1 where |ar| < 2, is
<
Ans. (b)
+
z32
(2)
9 z02 = z12 + z22 + z32 + 2 ( z12 + z22 + z33 )
3z02 = z12 + z22 + z32 .
Example 91: If w is an imaginary cube root of unity,
then value of the expression
1(2 – w) (2 – w2) + 2(3 – w) (3 – w2) + …
+ (n – 1) (n – w) (n – w2) is
(a)
1 2
n (n + 1)2 - n
4
(b)
1 2
n (n + 1)2 + n
4
(c)
1 2
n (n + 1) - n
4
(d)
1
n (n + 1)2 - n
4
Ans. (a)
n -1
=
r =1
r =0
(a) 0
(c) 4
Ans. (a)
z2 z3 + z2 z1 + z1 z3 =
From (1) and (2) we get
 ÈÎ(r + 1)3 - 1˘˚
Example 92: The number of complex numbers z such
1– z
«0
Thus,
1+ z
¤
S=
1 - ( x 2 + y 2 ) - 2 iy
As x ≥ 1, and y ≥ 0, we get
fi
Solution: rth term of the given expression is
r (r + 1 – w) (r + 1 –w2)
= (r + 1 – 1) (r + 1 – w) (r + 1 – w2)
= (r + 1)3 – 1
[\ (x – 1) (x – w) (x – w2) = x3 – 1]
Example 93: If a, b, c are integers, not all equal and
w (π1) is a cube root of unity, then minimum value of
|a + bw + cw 2| is
(a) 3
(c) 2
Ans. (b)
(b) 1
(d) 3
Solution: a + bw + cw 2
2
= (a + bw + cw 2 ) (a + bw + cw 2 )
= (a + bw + cw 2 ) (a + bw 2 + cw)
= a 2 + b 2 + c2 + (bc + ca + ab) (w + w2 )
= a 2 + b2 + c2 – bc – ca – ab
1
= [(b – c) 2 + (c – a)2 + (a – b) 2 ]
2
As a, b, c are integers not all equal, at least two of b – c,
c – a and a – b are of different from zero. Therefore minimum value of (b – c) 2 + (c – a) 2 + (a – b 2) is 2.
Thus, a + bw + cw 2
2
≥1
fi
|a + bw + cw 2| ≥ 1.
Example 94: The region of the argand plane defined by
|z – i| + |z + i| £ 4 is
(a) interior of an ellipse
(b) exterior of a circle
(c) interior and boundary of an ellipse
(d) none of these
Ans. (c)
2.32
Complete Mathematics—JEE Main
= |arg (B) – arg (A)|
= –BOA = –BOX – –AOX
= p/2 + (p/2 – q) – q = p – 2q
= 2 cos–1 (4/5)
Solution: Let A and B be the point 0 + i and 0 – i and
P(z) be any point satisfying |z – i| + |z + i| £ 4.
fi
PA + PB £ 4
Thus, P lies in the interior or on the boundary on the ellipse
with foci A and B and length of major axis = 4. See Fig. 2.51
Y
Y
A
P
C 25i
X
B
PA + PB = 4
B
15
15
O
25
A
q
Fig. 2.51
O
Example 95: If z1 and z2 are two non-zero complex
numbers such that |z1 + z2| = |z1| + |z2|, then arg (z1) – arg (z2)
is equal to
(a) – p
(b) – p/2
(c) p/2
(d) 0
Ans. (d)
Solution: Let z1 = r1(cos q1 + i sin q1) and
z2 = r2(cos q2 + i sin q2) where r1 = |z1|, r2 = |z2|, q1 =
arg (z1) and q2 = arg (z2).
We have
|z1 + z2|2 = r 12 + r22 + 2r1r2 cos (q1 – q2)
= (r1 + r2)2 + 2r1r2 {cos (q1 – q2) – 1}
Now, |z1 + z2| = |z1| + |z2|
¤ cos (q1 – q2) = 1 ¤ q1 – q2 = 0 ¤ q1 = q2.
Example 96: If |z – 25i| £ 15, then
|maximum arg (z) – minimum arg (z)| equals
–1
(a) 2 cos (3/5)
(c) p/2 + cos–1 (3/5)
Ans. (b)
–1
(b) 2 cos (4/5)
(d) sin–1 (3/5) – cos–1 (3/5)
Solution: If |z – 25i| £ 15, then z lies either in the
interior and or on the boundary of the circle with centre at
C (25i) and radius equal to 15.
From Fig. 2.52 it is clear that least argument is for point A
and the greatest argument is for point B.
OA
20
p
From right DOAC, cos Ê – q ˆ =
=
Ë2
¯ OC
25
fi
p/2 – q = cos–1 (4/5)
Now, for |z – 25i| £ 15
|maximum (arg z) – minimum (arg z)|
X
Fig. 2.52
Example 97: If |z| = 3, the area of the triangle whose
sides are z, wz and z + wz (where w is a complex cube root
of unity) is
(b) 3 3 /2
(a) 9 3 /4
(c) 5/2
(d) 8 3 /3
Ans. (a)
Solution: We have |z| = 3, |wz| = |w| |z| = (1) (3) = 3
and
|z + wz| = |(1 + w)z| = |(– w2)z|
= |– w2| |z| = (1) (3) = 3.
\ The given triangle is equilateral and its area is
3 2 9 3
|z| =
.
4
4
Example 98: The greatest and the least value of |z1 +
z2| if z1 = 24 + 7i and |z2| = 6 are respectively
(a) 31, 19
(b) 25, 19
(c) 31, 25
(d) none of these
Ans. (a)
Solution: Note that |z| = 6 represents a circle. As |z2|
= 6, |z1 + z2| = |z2 – (– 24 – 7i)| represent distance between a
point on the circle |z| = 6 and the point (– 24 – 7i).
|z1 + z2| will be greatest and least at points B and A which
are the end points of the diameter of the circle through C.
As OC = 25, CA = OC – OA= 25 – 6 = 19 and CB = OC +
OB = 25 + 6 = 31. See Fig. 2.53
Complex Numbers 2.33
1
£ 1 fi |z| 2 – |z| – 1 £ 0
z
Thus, |z| –
B
O
Fig. 2.53
|z2| = 6 fi z2 = 6e iq where q e R.
\
|z1 + z2|2 = |24 + 7i + 6 (cos q + i sin q)|2
= (24 + 6 cos q)2 + (7 + 6 sin q)2
= 576 + 36 cos2 q + 288 cos q + 49 +
36 sin2 q + 84 sin q
= 625 + 36 + 12 (24 cos q + 7 sin q)
= 661 + 12 (25) sin (q + a)
[put 7 = r cos a and 24 = r sin a]
= 661 + 300 sin (q + a)
Thus, greatest possible value of |z1 + z2|2 is 661 + 300 =
961 and the least possible value of |z1 + z2|2 is 361.
\ greatest and least possible values of |z1 + z2| are 31 and
19 respectively.
Example 99: If a, b are the roots of x2 + px + q = 0, and w
is a cube root of unity, then value of (wa + w2b) (w2a + wb) is
(b) 3q
(a) p2
(d) p2 – 3q
(c) p2 – 2q
Ans. (d)
Solution: We have a + b = – p, ab = q
(wa + w 2b ) (w2a + wb)
= w3 a2 + w4 ab + w2 ab + w3b 2
= a 2 + b2 + (w + w2) ab = a2 + b2 – ab
= (a + b)2 – 3ab = p2 – 3q
Example 100: Maximum distance from the origin of the
points z satisfying the relation |z + 1/z| = 1 is
5
(b) ( 5 – 1)/2
(d) (3 +
5 )/2
Ans. (a)
1
for otherwise, we
z
may interchange z and 1/z in the given equation.
We have
Solution: We may assume |z| ≥
|z| –
1
(1 –
2
5 ) £ |z| £
1
(1 +
2
5)
possible value of |z| is ( 5 + 1)/2.
Alternate Solution
(c) 3 –
fi
1
( 5 + 1).
2
1
i
Taking z = ( 5 + 1), we get z + = 1. Thus, maximum
z
2
C(-24 - 7i)
(a) ( 5 + 1)/2
|z| lies between the roots of |z|2 – |z| – 1 = 0
As z π 0, |z| > 0, therefore, 0 < |z| £
A
Now
fi
1
1
1
1
Ê 1ˆ
= z= z - - £ z - Á- ˜ = z + = 1
Ë z¯
z
z
z
z
Example 101: If |z1| = |z2| = |z3| = 1 and z1 + z2 + z3 = 0,
then area of the triangle whose vertices are z1, z2, z3 is
(a) 3 3 /4
(c) 1
Ans. (a)
(b) 3 /4
(d) 2
Solution: |z2 – z3|2 + |z2 + z3|2 = 2|z2|2 + 2|z3|2
fi
|z2 – z3|2 + |– z1|2 = 2(1) + 2(1)
[
fi
|z2 – z3| =
z1 + z2 + z3 = 0]
3
Similarly, |z3 – z1| = |z1 – z2| = 3
Thus, area of triangle with vertices z1, z2, z3 is
3
4
2
( 3)
=
3 3
4
Example 102: An equation of straight line joining the
complex numbers a and ib (where a, b e R and a, b π 0) is
Ê1 iˆ
Ê1 iˆ
(a) z Á - ˜ + z Á + ˜ = 2
Ë a b¯
Ë a b¯
(b) z(a – ib) + z (a + ib) = 2(a2 + b2)
(c) z(a + ib) + z (a – ib) = 2ab
(d) none of these
Ans. (a)
Solution: An equation of straight line joining a and ib is
z
z 1
a a 1 =0
ib -ib 1
fi z(a + ib) – z (a – ib) – 2iab = 0
Ê1 iˆ
Ê1 iˆ
fiz Á - ˜ + z Á + ˜ =2
Ë a b¯
Ë a b¯
Example 103: Two non-parallel lines meet the circle
|z| = r in the points a, b and c, d respectively. The point of
intersection of these lines is
2.34
Complete Mathematics—JEE Main
(a)
(b)
(c)
a -1 + b -1 + c -1 + d -1
z r2
r2
(b – a) + z (b – a) = (b2 – a 2)
ba
ab
fi
a -1b -1 + c -1d -1
ab + cd
a+b+c+d
za–1 b–1 +
fi
a -1 + b -1 - c -1 - d -1
r2
= a–1 + b–1
(2)
a
a -1b -1 - c -1d -1
(d) none of these
b
O
Ans. (c)
z
d
c
Solution: An equation of straight line passing through
A(a) and B(b) is
z z 1
a a 1 =0
b b 1
fi
z
z( a – b ) – z (a - b) + a b – a b = 0
Fig. 2.54
Similarly, equation of straight line joining c and d is
z c–1 d–1 +
(1)
z
r
2
= c–1 + d –1
(3)
Subtracting (3) from (2), we get
As a, b lie on |z| = r, we get
z(a–1 b–1 – c–1 d –1) = a–1 + b–1 – c–1 – d –1
|a| = |b| = r fi aa = bb = r 2.
Equation (1), now can be written as
fi
z=
Ê r2 r2 ˆ
a r2 b r2
z Á - ˜ - z (a - b) +
=0
b¯
b
a
Ë a
a -1 + b -1 - c -1 - d -1
a -1b -1 - c -1d -1
EXERCISE
Concept-based
Straight Objective Type Questions
1. The number of complex numbers z such that (1 + i)z
= i |z|
(a) 0
(b) 1
(c) 2
(d) infinite
2. Suppose a, b, c ΠR and C < 0. Let z = a + (b +
ic)2015 + (b – ic)2015, then
(a) Re (z) = 0
(b) Im (z) = 0
(c) Re (z) > 0, Im (z) < 0
(d) Re (z) < 0, Im (z) > 0
3. The number of solutions of z2 + |z| = 0 is
(a) 1
(b) 2
(c) 3
(d) infinite
(1 + i )z - 2
=k
4. The equation
(1 + i )z + 4
does not represent a circle when k is
(a) 2
(b) p
(c) e
(d) 1
5, then least value of z -
5. If |z|
(a) 5
(c) 8
(b) 24/5
(d) 8/3
i -1
is
2p
2p
i 1 - cos
+ sin
7
7
3p
(b)
28
19p
(d)
28
6. Principal argument of z =
p
28
17p
(c)
28
(a)
7. If x + iy =
(a) a2 + b2
(c)
a 2 + b2
c2 + d 2
1
is
z
(
)
a + ib
, then (x2 + y2)2 (c2 + d2) equals
c + id
(b)
(d)
a 2 + b2
a 2 + b2
c2 + d 2
Complex Numbers 2.35
8. Suppose z1, z2, z3 are three complex numbers, and
1 z1
1
D=
1 z2
4i
1 z2
then
(a) Re (D) = 0
(c) Re (D) ≥ 0
z1
z2 ,
z3
(a) 0, p
p 3p
,
(c)
2 2
3
11. If z is purely imaginary and Im (z) < 0, then
arg (i z) + arg (z) is equal to
(a) p
(c) p/2
(b) 0
(d) – p/2
12. The inequality a + ib > c + id is true when
(a) a > c, b > d > 0
(b) a > c, b = d = 0
(c) a > c, b = d > 0
(d) none of these
13. Let z ΠC be such that Re(z2) = 0, then
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
(b) (a, 0)
(d) (b, a)
(d)
(d) 0, 2p
16. If a > 0 and z|z| + az + 3i = 0, then z is
10. If w (π 1) is a cube root of unity, then the value
of tan [(w 2017 + w 2225) p – p/3]
1
1
(b)
(a) 3
3
(c) - 3
(b) p, –p
Ê z - 4ˆ p
15. If z = x + iy and 0 £ sin–1 Ë
£ then
2i ¯ 2
(a) x = 4, 0 £ y £ 2
(b) 0 £ x £ 4, 0 £ y £ 2
(c) x = 0, 0 £ y £ 2
(d) none of these
(b) Im (D) = 0
(d) Im (D) £ 0
9. If x, y, a, b Œ R, a π 0 and
(a + ib) (x + iy) = (a2 + b2) i,
then (x, y) equals
(a) (a, b)
(c) (0, b)
Êz ˆ
Êz ˆ
arg Á 1 ˜ + arg Á 2 ˜ equals
Ë z4 ¯
Ë z3 ¯
|Re(z)| + Im (z) = 0
|Re(z)| = |Im (z)|
Re(z) + |Im (z)| = 0
Re(z) = 0 or Im (z) = 0
0
purely imaginary
a positive real number
a negative real number
17. If z π 0 is a complex number such that Re(z) = 0,
then
(b) Im (z2) = 0
(a) Re (z2) = 0
2
2
(c) Re (z ) = Im (z ) (d) Im (z2) < 0
Ê kp ˆ
Ê kp ˆ
18. If zk = cos Ë ¯ + i sin Ë ¯ , then z1 z2 z3 z4 is equal
10
10
to
(a) –1
(b) 2
(c) –2
(d) 1
z +z
19. If |z1| = |z2| = 1, z1 z2 π –1 and z = 1 2 then
1 + z1z2
(a) z is a purely real number
(b) z is a purely imaginary number
(c) |z| = 1
(d) none of these
20. If z ΠC, then Re(z 2) = k2, k > 0, represents
(a) an ellipse
(c) a circle
14. If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then
(b) a parabola
(d) a hyperbola
LEVEL 1
Straight Objective Type Questions
21. If w π 1 is a cube root of unity, then 1, w, w2
(a)
(b)
(c)
(d)
are vertices of an equilateral triangle
lie on a straight line
lie on a circle of radius 3/2
none of these
22. If a, b, g are the cube roots of p, p < 0, then for
any x, y and z which does not make denominator
xa + yb + zg
zero, the expression
equals
x b + yg + za
(a) w, 1
(c) w2, 1
(b) w, w2
(d) 1, w, w2
23. ABCD is a rhombus, its diagonal AC and BD intersect at the point M and satisfy BD = 2AC. If the
points D and M are the complex numbers 1 + i and
2 – i respectively, then A represent the complex
number.
2.36
Complete Mathematics—JEE Main
(a) 3 –
1
3
i, 1 + i
2
2
1
3
(b) 3 + i, 1 + i
2
2
(c) 3 –
1
3
i, 1 – i
2
2
1
3
(d) 3 + i, 1 - i
2
2
24. Let a and b be the roots of the equation x 2 + x +
1 = 0. The equation whose roots are a 19, b 7 is
(a) x 2 – x – 1 = 0
(c) x 2 + x – 1 = 0
(b) x2 – x + 1 = 0
(d) x2 + x + 1 = 0
25. If w is an imaginary cube root of unity, then the
value of
p
sin (w 10 + w 23 ) p 4
is
{
}
3
2
(a) –
(b) –
1
2
1
3
(c)
(d)
2
2
26. If w ( π 1) is a cube root of unity and (1 + w)2017 =
A + Bw. Then A and B are respectively the numbers
(a) 0,1
(c) 1, 0
(b) 1, 1
(d) – 1, 1
27. If w (π 1) is a cube root of unity, then
1+ i +w
1
2
w
2
1– i
–1
w2 – 1
–i –i + w – 1
–1
equals
(a) 0
(c) i
(b) 1
(d) w
28. If w (π 1), is a cube root of unity, then value of
7
(1 + w – w 2 ) equals
(a) 128w
(c) 128w2
(b) – 128w
(d) – 128w 2
6i –3i 1
29. If D = 4 3i –1 = x + iy , then
20 3
i
(a) x = 3, y = 1
(b) x = 1, y = 3
(c) x = 0, y = 3
(d) x = 0, y = 0
30. If arg(z) < 0, then arg (– z) – arg (z) equals
(a) p
(b) – p
p
p
(d)
2
2
31. If z1, z2, z3 are complex numbers such that
(c) –
z1 = z2 = z3 =
1
1
1
+
+
=1,
z1 z2 z3
then |z1 + z2 + z3| is
(a) equal to 1
(b) less than 1
(c) greater than 3
(d) equal to 3
32. Let z1 and z2 be nth roots of unity which subtend
a right angle at the origin. Then n must be of the
form
(a) 4k + 1
(b) 4k + 2
(c) 4k + 3
(d) 4k
33. The complex numbers z1, z2 and z3 satisfying
z1 – z3
1
= (1 – 3i ) are vertices of a triangle
z2 – z3
2
which is
(a) of area zero
(c) equilateral
34. Let w = –
(b) right-angled isosceles
(d) obtuse-angle isosceles
1
3
+
i . Then the value of the determi2
2
nant
1
1
D = 1 –1–w
1
w2
(a) 3w
(c) 3w2
1
2
w2
is
w4
(b) 3w (w – 1)
(d) 3w (1 – w)
35. The inequality |z – i| < |z + i| represents the region
(a) Re(z) > 0
(c) Im(z) > 0
(b) Re(z) < 0
(d) Im(z) < 0
36. If iz 3 + z2 – z + i = 0, then
(a) Re z = 0
(c) |z| = 1
(b) Im z = 0
(d) none of these.
1
, q π 2np, n Œ I, then
1 – cos q + 2i sin q
maximum value of x is
37. If x + iy =
(a) 1
(b) 2
1
1
(d)
2
3
38. The equation z 3 = z has
(c)
(a)
(b)
(c)
(d)
no solution
two solutions
five solutions
infinite number of solutions
39. If z = 5 + t + i 25 – t 2 , (– 5 £ t £ 5), then locus
of z is a curve which passes through
(a) 5 + 0i
(b) – 2 + 3i
(c) 2 + 4i
(d) – 2 – 3i
40. If w π 1 is a cube root of unity and satisfies
1
1
1
+
+
= 2w 2
a +w b+w c +w
Complex Numbers 2.37
1
and
a +w
2
+
1
2
1
+
2
= 2w,
b+w
c +w
1
1
1
+
+
is
then the value of
a +1 b +1
c +1
(a) 2
(b) – 2
(d) none of these
(c) – 1 + w2
41. If |z – i Re(z)| = |z – Im(z)|, then z lies on
(a) Re (z) = 2
(b) Im(z) = 2,
(c) Re(z) + Im(z) = 2 (d) none of these
42. If w is a complex cube root of unity, then value of
expression
È
cos Í{(1 – w) (1 – w2) +
+
Î
p ˘
(12 – w) (12 – w2)}
370 ˙˚
(a) – 1
(b) 0
(c) 1
(d)
3 2
43. If roots of the equation z + az + b = 0 are purely
imaginary then
(b – b )2 + (a + a ) ( a b + a b ) = 0
(b – b )2 + (a – a )2 = 0
(b + b )2 – (a – a )2 = 0
none of these
44. The system of equations |z + 1 – i| =
= 3 has
(a) no solution
(b) one solution
(c) two solutions
(d) infinite number of solutions
2 and |z|
45. If 8i z3 + 12z2 – 18z + 27i = 0, then
2
3
3
(c) |z| = 1
(d) |z| =
4
46. If the complex number z lies on the boundary of the
circle of radius 5 and centre at 4, then the greatest
value of |z + 1|
(a) |z| =
3
2
(a) 4
(c) 10
47. If x + iy =
reduces to
(a) 2
(c) 4
(b) 2z1 + z2 + z3 = 0
(d) 4z1 + z2 + z3 = 0
49. Suppose that three points z1, z2, z3 are connected
by the relation az1 + bz2 + cz3 = 0, where a + b +
c = 0, then the points are
(a) vertices of a right triangle
(b) vertices of an isosceles triangle
(c) vertices of an equilateral triangle
(d) collinear
z -1
is purely imaginary,
50. If the complex number
z +1
then
(a) |z| = 1
(c) |z| > 1
(b) |z| < 1
(d) |z| ≥ 2
51. If z is a complex number such that –p/2 £
arg z £ p/2, then which of the following inequality
is true.
2
(a)
(b)
(c)
(d)
(a) z1 + z2 + z3 = 0
(c) z1 + z2 + 4z3 = 0
(b) |z| =
(b) 5
(d) 9
3
, then 4x – x2 – y2
cos q + i sin q + 2
(b) 3
(d) 5
48. Suppose z1, z2, z3 represent the vertices A, B, and
C respectively of a DABC with centroid at G. If the
mid point of AG is the origin, then
(a)
(b)
(c)
(d)
| z – z | £ | z | | arg (z) – arg ( z )|
| z – z | £ | arg (z) – arg ( z )|
| z – z | > | z | | arg (z) – arg ( z )|
none of these
52. If |w | = 1, then the set of points
1
is satisfies
z=w+
w
(a) |Re (z)| £ 2
(b) |z| £ 1
(c) |z| = 1
(d) |Im (z)| ≥ 2
53. Number of complex numbers satisfying |z| = 1 and
z z
+
= 1, is
z z
(a) 0
(b) 2
(c) 4
(d) 8
54. If z1, z2 are two complex numbers such that |z1| =
|z2| = 2 and |z1 + z2| = 3 , then |z1 – z2| equals:
(a) 2 2
(c) 3
(b)
5
(d) 2 - 2
55. Let z1, z2, z3 be three non-zero complex numbers
such that z1 z2 = z2 z3 = z3 z1 , then z1, z2, z3
(a)
(b)
(c)
(d)
are vertices of an equilateral triangle
are vertices of an isosceles triangle
lie on a straight line
none of these
56. If |z1| = |z2| = |z3| = 1, then value of |z2 – z3|2 +
|z3 – z1|2 + |z1 – z2|2 cannot exceed
(a) 3
(c) 9
(b) 6
(d) 12
57. Let z1, z2, z3 be three complex number such that
z1 + z2 + z3 = 0 and |z1| = |z2| = |z3| = 1, then
z12 + 2 z22 + z32 equals
2.38
Complete Mathematics—JEE Main
(a) 1
(c) 3
(b) 2
(d) 4
58. Let z1, z2, z3 be three complex numbers such that
|z1| = |z2| = |z3| = 1 and
Ê1
1
1ˆ
+ ˜ , then |z| cannot
z = ( z1 + z2 + z3 ) Á +
Ë z1 z2 z3 ¯
exceed
(a) 1
(c) 6
(b) 3
(d) 9
59. Suppose z is a complex number such that z π –1,
z(1 - z )
|z| = 1, and arg (z) = q. Let w =
, then
z (1 + z )
Re (w) is equal to
(a) 1 + cos (q/2)
(c) –2 sin2 (q/2)
(b) 1 – sin (q/2)
(d) 2 cos2 (q/2)
Ê 1 + z2 ˆ
, where z is any non-zero com60. Let a = Im Á
Ë 2iz ˜¯
plex number. Then the set
A = {a : |z| = 1 and z π ± 1}
is equal to
(a) (–1, 1)
(c) [0, 1)
(b) [–1, 1]
(d) (–1, 0]
61. Number of complex numbers such that
|z| = 1 and z = 1 – 2z is
(a) 0
(b) 1
(c) 2
(d) infinite
(a) 5
(b) –5
(c) 2 3 i
(d) -2 3 i
64. Suppose a < 0 and z1, z2, z3, z4 be the fourth roots
2
2
2
2
of a. Then z1 + z2 + z3 + z4 is equal
(b) |a| – a
(d) a2
65. Suppose arg (z) = – 5p/13, then arg
(a) – 5p/13
(c) p
68. Im
(b) Re (z) < 1
(d) Im (z) < 1
Ê 2 z + 1ˆ
= 5 represents
Ë iz + 1 ¯
(a) a circle
(c) a parabola
(b) a straight line
(d) an ellipse
69. Let z1, z2 be two complex numbers such that Im (z1
+ z2) = 0 and Im (z1 z2) = 0 then
(a) z1 = – z2
(c) z1 = z2
70. If n ΠN, then
(b) z1 = z2
(d) none of these
(1 + i )n
(1 - i )n - 2
(a) i n +1
(c) i n +2
is equal to
(b) –2i n +1
(d) –2i n +2
71. Let w π 1, be a cube
be defined by
f(n) = 1 + w n + w 2n,
(a) {0}
(c) {0, 1, 3}
root of unity, and f : I Æ C
then range of f is
(b) {0, 3}
(d) {0, 1}
1
= 2 cosq, z ΠC then z2n Р2zn cos (nq)
z
is equal to
(a) 1
(c) –1
(b) 0
(d) –n
73. If w π 1 is a cube root of unity, then
60
(d) 0
63. If z = i (1 + 3 ) , then z4 + 2z3 + 4z2 + 5 is equal to
(a) – a2
(c) a + |a|
(a) Re (z) > 1
(c) Im (z) > 1
72. If z +
62. Let z1, z2 be two complex numbers such that z1 π 0
2iz1 + 5z2
is equal to
and z2/z1 is purely real, then
2iz1 - 5z2
(a) 3
(b) 2
(c) 1
67. If z Œ C – {0, –2} is such that
log(1/7) |z – 2| > log(1/7) |z|
then
z=
30
Âw k - ’w k
k =1
is equal to
k =1
(a) 0
(c) w 2
(b) w
(d) –1
74. Let g(x) and h(x) be two polynomials with real
coefficients. If P(x) = g(x3) + xh(x3) is divisible by
x2 + x + 1, then
(a) g(1) = 0, h(1) = 1
Ê z+z ˆ
is
Ë 1 + zz ¯
(b) 5p/13
(d) 0
66. The number of values of q Π(0, p], such that (cos q +
i sin q) (cos 3q + i sin 3q) (cos 5q + i sin 5q) (cos 7q
+ i sin 7q) (cos 9q + i sin 9q) = –1 is
(a) 11
(b) 13
(c) 14
(d) 16
(b) g(1) = 1, h(1) = 0
(c) g(1) = 0, h(1) = 0
(d) g(1) = 1, h(1) = 0
75. If x2 – x + 1 divides the polynomial xn+1 – xn + 1,
then n must be of the form
(a) 3k + 1
(b) 6k + 1
(c) 6k – 1
(d) 3k – 1
Complex Numbers 2.39
Assertion-Reason Type Questions
76. Statement-1: z2 + z |z Р1| + |z Р1|2 = 0, z ΠC
has no solution in C.
Statement-2: z2 + az + a2 = 0, z ΠC, a > 0 has
no solution in C.
Statement-2: A square matrix A with complex
entries is non-singular if and only if |A| π 0
81. Statement-1: If |z1| |z2| = 1, then
|z1 – z2| =
77. Let z ΠC satisfy the relations
|z| = 1 and z = 2i + z
Statement-2: | z | = | z | " z ΠC.
Statement-1: z is purely imaginary.
82. Let f(z) be a polynomial in z with complex coefficients. Suppose, when f(z) is divided by z – i, the
remainder is i and when f(z) is divided by z + i,
the remainder is 1 + i.
Statement-2: |arg (z)| < p/6.
Ê8p ˆ
Ê3p ˆ
78. Let a = cos Ë ¯ + i sin Ë ¯
11
11
Statement-1: Re (a + a2 + a3 + a4 + a5) = –
10
Statement-2:
Statement-1: The remainder when f(z) is divided
1
by z2 + 1 is (iz + 1) + 1
2
Statement-2: If r(z) is remainder when f(z) is divided by p(z), then
r(z) = 0 or deg (r(z)) < deg (p(z)).
1
2
 ak = 0
k =0
79. Let z = cos q + i sin q. q Π(0, p), a = p/60
15
and
f(q) =
83. Let a, b Œ C, a π 0. Suppose z1, z2, z3 are the roots
of the equation (z + ab)3 = a3.
 sin[(2k - 1)q ] .
k =1
Statement-1: f(a) =
1
2 sina
3
1 1
z1 z2
Statement-1: Length of a side of the triangle with ver-
Statement-2: z + z + … + z
29
=
tices z1, z2 and z3 is
z(1 - z 30 )
Statement-2: Roots of z3 = a3 lie on a circle of
radius |a|.
2
1- z
80. Let w π 1 be a cube root of unity and S be the set
of all non-singular matrices of the form
È 1 a b˘
Í w 1 c˙
˙
Í
ÍÎw 2 w 1 ˙˚
3|a|
84. Statement-1: The equation z2 – z + p = 0.
has no solution in the unit disc |z| < 1.
Statement-2: |z1 + z2| £ |z1| + |z2| " z1, z2 Œ C.
85. Let z1 = r1eiq, z2 = r2eij where q, j Œ (–p, p].
where each of a, b and c is either w or w 2.
Statement-1: |z1 – z2|2 < (r1 –r2)2 + (q – f)2
Statement-1: S contains exactly two distinct
matrices.
Statement-2: |sin q | < | q | " q ΠR.
LEVEL 2
Straight Objective Type Questions
86. If the complex numbers z1, z2, z3, are the vertices
of a parallelogram ABCD, then the fourth vertex is
(a)
1
(z1 + z2)
2
1
(c)
(z1 + z2 + z3)
3
(b)
1
(z1 + z2 – z3 – z4)
4
(d) z1 + z3 – z2
87. If a, b and c are three integers such that at least
two of them are unequal and w (π 1) is a cube root
of unity, then the least value of the expression |a +
bw + cw 2 | is
(a) 0
(c)
(b) 1
3
2
(d)
1
2
2.40
Complete Mathematics—JEE Main
88. The shaded region in Fig. 2.55 is given by
95. If z1, z2, z3 ΠC are distinct and are such that |z1|
= |z2| = |z3| and z1 + z2 + z3 = 0, then z1, z2, z3
(a)
(b)
(c)
(d)
are vertices of a right triangle
an equilateral triangle
an obtuse angled triangle
none of these
96. If w = cos(p/n) + i sin(p/n), then value of 1 + w +
+w n – 1 is
w2 +
(a)
(b)
(c)
(d)
Fig. 2.55
(a)
(b)
(c)
(d)
{
{
{
{
z : z - 1 < 2, arg ( z + 1) <
p
2
p
z : z + 1 < 2, arg ( z + 1) <
2
z : z - 1 > 2, arg ( z - 1) <
p
4
z : z + 1 > 2, arg ( z + 1) <
p
4
}
}
}
}
89. Let w = a + ib, b π 0 and z π 1. If
97. Let z1, z2 be two non-zero complex numbers such
z
z
that |z1 + z2| = |z1 – z2|, then 1 + 2 equals
z1 z2
(a) 0
(c) – 1
w – wz
is
1– z
purely real, then the set of value of z is
(a) {z : |z| = 1}
(b) {z : z = z}
(c) {z : |z| π 1}
(d) {z : |z| = 1, z π 1}
90. If z and w are non-zero complex numbers such that
–
–
zw = | z |2, then | z – z | + | w + w | = 4 represents a
(a) rectangle
(b) square
(c) rhombus
(d) trapezium
91. Let z be a complex number such that z = 1 – t +
i t 2 + t + 2 , where t ΠR, then locus of z on the
Argand plane is
(a) a parabola
(b) an ellipse
(c) a hyperbola
(d) a pair of straight line.
92. Let z1, z2, ΠC and a, b > 0 be such that
az – bz2
lies
a|z1| = b|z2|, then w = 1
az1 + bz2
(a) in the 1st quadrant (b) in the 3rd quadrant
(c) on the real axis
(d) on the imaginary axis
93. If z ΠC, then least value of the expression |z| +
|1 – z| + |z – 2| is
(a) 1
(c) 2
(b) 3/2
(d) cannot be determined
94. If k > 0, k π 1, and z1, z2 Œ C, then
represents
(a) a circle
(c) a parabola
(b) an ellipse
(d) a hyperbola
1+i
1 + itan(p/2n)
1 + icot(p/2n)
none of these
z – z1
= k
z – z2
(b) 1
(d) none of these
98. If |z1| = 2 and (1 – i)z2 + (1 + i) z2 = 8 2 , then
(a) minimum value of |z1 – z2| is 1
(b) minimum value of |z1 – z2| is 2
(c) maximum value of |z1 – z2| is 8
(d) maximum value of |z1 – z2| is 4
99. If z1 lies on |z| = r, then equation of tangent at z1
is
(a)
z
z
+ =2
z1 z1
(b)
z
z
+ =r
z1 z1
(c)
z
z
+ =2
z1 z1
(d)
z
z
+ =r
z1 z1
100. If z ΠC, then minimum value of |z Р2 + 3i| +
|z – 1 + i| is
(a)
(c)
5
13 – 2
(b) 2 5
(d) 0
101. If a > 0 and |z – a 2| + |z – 2a| = 3 represents an
ellipse then a lies in
(a) (0, 3)
(b) (0, •)
(c) (1, 3)
(d) (3, •)
102. If the points A(z), B(– z) and C(1 – z) are the vertices
of an equilateral triangle, then value of z is
1
(1 ± i )
2
1
1
(c)
(d) (1 ± 3i )
(
1 ± 3i )
4
3
103. If |z + 1| + |z – 3| £ 10, then the range of values
of |z – 7| is
(a) 1 ±
i 3
2
(a) [0, 10]
(c) [2, 12]
(b)
(b) [3, 13]
(d) [7, 9]
Complex Numbers 2.41
104. If w π 1 is a cube root of unity and |z – 1| 2 + 2|z
2
– w | 2 = 3|z – w 2| then z lies on
(a)
(b)
(c)
(d)
a straight line
a parabola
an ellipse
a rectangular hyperbola
105. If x > 0, the least value of n ΠN such that
n
2
2
Ê1 + iˆ
–1 Ê 1 + x ˆ is
˜
ÁË 1 – i ˜¯ = sin ÁË
p
2x ¯
(a) 2
(c) 8
(b) 4
(d) 32
Previous Years' AIEEE/JEE Main Questions
1. z and w are two non-zero complex numbers such
that |z| = |w| and Arg z + Arg w = p, then z equals
(b)
(d) – w
(a)
(c) w
[2002]
(c)
p
4
(d)
Ê x yˆ
Á + ˜
p + q Ë p q¯
1
2
(b) Re(z) < 0
(d) Re(z) > 2
[2002]
3. The locus of the centre of a circle which touches the
circles |z – z1| = a and |z – z2| = b, a π b, externally
is
(a) an ellipse
(c) a circle
(b) a hyperbola
(d) a pair of straight lines
[2002]
x
1 + iˆ
4. If ÊÁ
= 1 , then
Ë 1 – i ˜¯
(a)
(b)
(c)
(d)
x = 2n, where n is any positive integer
x = 4n + 1, where n is any positive integer
x = 2n + 1, where n is any positive integer
x = 4n, where n is any positive integer
[2003]
5. If z and w are two non-zero complex numbers such
p
, then z w
that |zw| = 1, and Arg(z) – Arg(w) =
2
is equal to
(a) – 1
(c) – i
(b) i
(d) 1
[2003]
6. Let z1 and z2 be two roots of the equation z2 + az
+ b = 0, z being complex. If the origin, z1 and z2
form an equilateral triangle, then
(a) a2 = 2b
(c) a2 = 4b
(b) a2 = 3b
(d) a2 = b
[2003]
7. Let z, w be two complex numbers such that
z + i w = 0 and arg(zw) = p, then arg(z) equals
(a)
3p
4
(b)
[2004]
8. If z = x – iy and z 1/3 = p + iq, then
2. If |z – 4| < |z – 2|, then
(a) Re(z) > 0
(c) Re(z) > 3
5p
4
p
2
2
is equal to
(a) 2
(c) 1
2
(b) – 1
(d) – 2
[2004]
2
9. If |z – 1| = |z| + 1, then z lies on
(a) a circle
(c) the real axis
(b) the imaginary axis
(d) an ellipse
[2004]
10. If the cube roots of unity 1, w, w2, then the roots
of the equation (x – 1) 3 + 8 = 0, are
(a)
(b)
(c)
(d)
– 1, 1 – 2w, 1 – 2w2
– 1, 1 + 2w, 1 + 2w2
– 1, – 1 + 2w, – 1 – 2w2
– 1, – 1, – 1
[2005]
z
and |w| = 1, then z lies on
1
z– i
3
(a) straight line
(b) a parabola
(c) an ellipse
(d) a circle
[2005]
11. If w =
12. If z1 and z2 are two non-zero complex numbers such
that |z1 + z2| = |z1| + |z2|, then arg z1 – arg z2 is
equal to
p
(a) 0
(b) –
2
(c)
p
2
(d) p
10
13. The value of
 ÊËsin
k =1
(a) – i
(c) 1
[2005]
2 kp
2 kp ˆ
is
+ i cos
11
11 ¯
(b) i
(d) – 1
[2006]
2.42
Complete Mathematics—JEE Main
14. If z 2 + z + 1 = 0, where z is a complex number,
then value of
2
1 ˆ2
1ˆ 2 Ê 2 1 ˆ
Ê
Ê
+ Á z + 2 ˜ + Á z3 + 3 ˜ +
z+
Ë
Ë
Ë
z¯
z ¯
z ¯
22. If z is a complex number of unit modulus and argu1+ zˆ
ment q, then arg Ê
equals
Ë1+ z ¯
p
(a)
-q
(b) q
2
(c) p – q
(d) –q
[2013]
+
2
Ê 6 1ˆ
ÁË z + 6 ˜¯ is
z
(a) 12
(c) 54
(b) 18
(d) 6
[2006]
15. If |z + 4| £ 3, then the maximum value of |z + 1|
is
(a) 4
(c) 6
(b) 10
(d) 0
–1
i –1
(b)
1
i +1
(c)
–1
i +1
(d)
1
i –1
(c)
(b) 2 +
3 +1
(d)
2
5 +1
[2009]
(b) •
(d) 1
[2010]
(b) (0,1)
(d) (1, 0)
[2011]
20. Let a, b be real and z be a complex number. If
z2 + az + b = 0 has two distinct roots on the line
Re (z) = 1, then it is necessary that
(a) b Œ (1, •)
(c) b (–1, 0)
[2013, online]
Ê 1 + z2 ˆ
24. Let a = Im Á
, where z is any non-zero comË 2iz ˜¯
plex number. The set A = {a : |z| = 1 and z π ±1}
is equal to:
(a) (–1, 1)
(c) [0, 1)
(b) [–1, 1]
(d) (–1, 0]
[2013, online]
25 . Let z ΠC satisfy |z| = 1 and z = 1 Рz .
19. If w (π1) is a cube root of unity and (1 + w)7
= A + Bw, there (A, B)
(a) (–1, 1)
(c) (1, 1)
(b) 5
(d) 1
†
18. The number of complex numbers z such that
|z – 1| = |z + 1| = |z – i| equals
(a) 2
(c) 0
(a) 2
(c) 3
[2008]
4
= 2, then the maximum value of |z| is
17. If z –
z
equal to
(a) 1
23. If z1 π 0 and z2 be two complex numbers such that
2 z1 + 3z2
z2/z1 is purely imaginary number, then
2 z1 - 3z2
is equal to:
[2007]
1
16. The conjugate of a complex number is
. Then
i
–
1
that number is
(a)
(d) either on the real axis or on a circle passing
through the origin.
[2012]
(b) b Π(0, 1)
(d) |b| = 1
[2011]
z2
is real, then the point represented
z -1
by the complex number z lies:
21. If z π 1 and
(a) on a circle with centre at the origin.
(b) either on the real axis or on a circle not passing
through the origin.
(c) on the imaginary-axis.
Statement-1: z is a real number
Statement-2: Principal argument of z is ± p/3.
[2013, online]
26. If a complex number z satisfies z +
= 0, then |z| is equal to:
(a) 2
(c) 5
(b) 3
(d) 1
2 |z + 1| + i
[2013, online]
27. If z is a complex number such that |z| ≥ 2, then
1
minimum value of z + :
2
(a)
(b)
(c)
(d)
is strictly greater than 5/2
is strictly greater that 3/2 but less than 5/2
is equal to 5/2
lies in the interval (1, 2)
[2014]
28. Let w (Im w π 0) be a complex number. Then the
set of all complex numbers z satisfying the equation
w – wz = k (1 – z), for some real number k, is:
(a) {z : |z| = 1}
(c) {z : z π 1}
(b) {z : z = z}
(d) {z : |z| = 1, z π 1}
[2014, online]
29†. If z1, z2 and z3, z4 are 2 pairs of complex conjugate
numbers and z1, z3 œ R, then
Êz ˆ
Êz ˆ
arg Á 1 ˜ + arg Á 2 ˜ equals:
Ë z4 ¯
Ë z3 ¯
† Slightly modified version.
Complex Numbers 2.43
(a) 0
(b)
p
2
(a)
3p
(d) p
[2014, online]
2
z-i
30. Let z π –i be any complex number such that
z+i
1
is a purely imaginary number. Then z + is:
z
(a) 0
(b) any non-zero real number other than 1
(c) any non-zero real number
(d) a purely imaginary number
[2014, online]
(c)
31. For all complex numbers z of the form 1 + ia,
a ΠR if z2 = x + iy, then
(a) y2 – 4x + 2 = 0
(c) y2 – 4x + 4 = 0
(b) y2 + 4x – 4 = 0
(d) y2 + 4x + 2 = 0
[2014, online]
32. A complex number z is said to be unimodular if |z|
= 1. Suppose z1 and z2 are complex numbers such
z - 2 z2
is unimodular and z2 is not unimoduthat 1
2 - z1 z2
lar. Then the point z1 lies on a:
(a)
(b)
(c)
(d)
straight line parallel to the x-axis.
straight line parallel to the y-axis.
circle of radius 2.
circle of radius 2 .
17
(b)
3
2
(d)
2
34. If z is a non-real complex
Im z 5
mum value of
is:
(Im z )5
(a) – 1
(b)
(c) – 4
(d)
(c)
2
5
2
[2015, online]
2
number, then the mini-
–2
–5
[2015, online]
2 + 3i sin q
is purely
35. A value of q for which z =
1 - 2i sin q
imaginary, is
p
p
(a)
(b)
3
6
Ê 3ˆ
Ê 1 ˆ
(c) sin 1 Á ˜
(d) sin 1 Á ˜
[2016]
Ë 3¯
Ë 4 ¯
36. The point represented by 2 + i in the Argand plane
moves 1 unit eastwards, then 2 units northwards and
finally from there 2 2 units in the south-westwards
direction. Then its new position in the Argand plane
is at the point represented by:
(a) 1 + i
(c) –2– 2i
(b) 2 + 2i
(d) –1 – i
[2016, online]
37. Let z = 1 + ai be a complex number, a > 0, such
that z3 is a real number. Then the sum 1 + z + z2
+ … + z11 is equal to:
[2015]
33. The largest value of r for which the region represented by the set {w Œ C: |w – 4 – i| £ r} is
contained in the region represented by the set {z
ŒC: |z – 1| £ |z + i|}, is equal to:
(a) 1365 3 i
(b) – 1365 3 i
(c) –1250 3 i
(d) 1250 3 i
[2016, online]
Previous Years’ B-Architecture Entrance
Examination Questions
1.
5 + i sin q
is a real number when
5 - 3i sin q
(a) q = p/4
(b) q = –p
(c) q = –p/2
(d) q = p/2
(a) 1
(c) 3
[2006]
2. Two points P and Q in the Argand diagram represent complex numbers z and 3z + 2 + i. If P
moves arround the circle with centre at the origin
and radius 2, then Q moves on the circle, whose
centre and radius are
(a) –2 – i, 6
(c) 2 + i, 6
(b) 2 – i, 3
(d) 2 + i, 3
[2007]
3. Let z be a complex number such that |z| = 2, then
2
is
maximum possible value of z +
z
(b) 2
(d) 4
Ê 1
3ˆ
4. If i = -1 , then 4 + 3 ÁË - + i ˜¯
2
2
is equal to
[2008]
127
Ê 1
3ˆ
+ 5 ÁË - + i ˜¯
2
2
(a) 4 3 i
(b) 2 3 i
(c) 1 - 3 i
(d) 1 + 3 i
124
[2009]
5. The real part of a complex number z having minimum
principal argument and satisfying |z – 5i| £ 1 is
2
6
(a)
(b) 0
5
2
1
6
(c)
(d) [2010]
5
5
2.44
Complete Mathematics—JEE Main
6. Area of a triangle with vertices given by z, iz,
z + iz, where z is a complex number, is
1 2
z
(a) 0
(b)
2
(c) | z |2
(d) 2| z |2
[2011]
7. Two circles in the complex plane are
C1 : |z – i| = 2
C2 : |z – 1 – 2i| = 4
then
(a) C1 and C2 touch each other
(b) C1 and C2 intersect at two distinct points
(c) C1 lies within C2
(d) C2 lies within C1
[2012]
8. If z = i (i + 2 ) , then value of z4 + 4z3 + 6z2 + 4z
is
(a) –5
(c) 6
(b) 3
(d) – 9
[2013]
9. Suppose z is a complex number such that z π –1,
z(1 - z )
, then Re(w)
|z| = 1 and arg(z) = q. Let w =
z (1 + z )
is equal to
Êq ˆ
(a) 1 + cos Ë ¯
2
Êq ˆ
(b) 1 - sin Ë ¯
2
Êq ˆ
Êq ˆ
(c) -2 sin 2 Ë ¯
(d) 2 cos2 Ë ¯
2
2
10. If |z1| = |z2| = |z3| = 1
and z1 + z2 + z3 =
[2014]
Level 1
21.
25.
29.
33.
37.
41.
45.
49.
53.
57.
61.
65.
69.
73.
77.
81.
85.
(a)
(c)
(d)
(c)
(c)
(d)
(a)
(d)
(d)
(a)
(a)
(c)
(d)
(d)
(c)
(b)
(a)
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
62.
66.
70.
74.
78.
82.
(b)
(b)
(a)
(b)
(c)
(c)
(c)
(a)
(b)
(d)
(c)
(b)
(b)
(c)
(a)
(a)
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
63.
67.
71.
75.
79.
83.
(c)
(a)
(a)
(c)
(c)
(a)
(b)
(a)
(a)
(c)
(a)
(a)
(b)
(b)
(a)
(b)
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
64.
68.
72.
76.
80.
84.
(d)
(d)
(d)
(c)
(a)
(a)
(d)
(a)
(c)
(a)
(c)
(a)
(c)
(c)
(a)
(a)
Level 2
86. (d)
87. (b)
88. (d)
89. (d)
90. (b)
91. (c)
92. (d)
93. (c)
94. (a)
95. (b)
96. (c)
97. (a)
98. (b)
99. (a)
100. (a)
101. (a)
102. (c)
103. (b)
104. (a)
105. (b)
Previous Years’ AIEEE/JEE Main Questions
2 + i, then the number
z1 z2 + z2 z3 + z3 z1 is:
1. (b)
2. (c)
3. (b)
4. (d)
(a)
(b)
(c)
(d)
5. (c)
6. (b)
7. (a)
8. (d)
9. (b)
10. (a)
11. (a)
12. (a)
[2015]
13. (a)
14. (a)
15. (c)
16. (c)
11. Let S = {z ΠC: z(iz1 Р1) = z1 + 1, [z1] < 1}. Then,
for all z ŒS, which one of the following is always true?
17. (d)
18. (d)
19. (c)
20. (a)
21. (d)
22. (b)
23. (d)
24. (a)
25. (d)
26. (c)
27. (d)
28. (d)
29. (a)
30. (no answer)
32. (c)
33. (d)
36. (a)
37. (a)
a positive real number
a negative real number
always zero
a purely imaginary number
(a) Re z + Im z < 0
(c) Re z – Im z > 1
(b) Re z < 0
(d) Re z – Im z < 0 [2016]
Answers
Concept-based
1.
5.
9.
13.
17.
(b)
(b)
(d)
(b)
(b)
2.
6.
10.
14.
18.
(b)
(c)
(c)
(d)
(a)
3.
7.
11.
15.
19.
(c)
(a)
(c)
(a)
(a)
4.
8.
12.
16.
20.
(d)
(b)
(b)
(b)
(d)
31. (b)
34. (c)
35. (d)
Previous Years’ B-Architecture Entrance
Examination Questions
1. (b)
2. (c)
3. (c)
4. (a)
5. (a)
6. (b)
7. (c)
8. (b)
9. (c)
10. (c)
11. (*)
Complex Numbers 2.45
Hints and Solutions
\
z=
Concept-based
1. |(1 + i)z| = | i | z ||
fi
fi
2 |z| = |z|
|z| = 0
fi
fi
=
( 2 – 1) |z| = 0
z=0
2. z = a + (b + ic)2015 + (b – ic)2015
= a + (b – ic)
\
+ (b + ic)
2015
fi
fi | x + iy|2 |c + id| = |a + ib|
fi (x2 + y2)
2
we get z = –1 fi z = ± i
2
(1 + i )
=k
4
z+
(1 + i )
c 2 + d 2 = a 2 + b2
1 z1
1
8. D =
1 z2
- 4i
1 z3
4. We can write the equation as
z-
(1)
z1
z2 = D
z3
fi D is purely real fi Im(D) = 0
9. (a + ib) (x + iy) = (a + ib) (a – ib) i
As a π 0
2
12 - i 2
=
=1–i
1+ i
1+ i
4
and
= 2(1 – i)
1+ i
Therefore, (1) can be written as
But
x + iy = (a – ib) i = b + ia fi (x, y) = (b, a)
10. w 2017 = (w 3)672 w = w
and w 2225 = (w 3)741 w 2 = w 2
\ tan [(w 2017 + w 2225)p – p/3]
(2)
This will not represent a circle if k = 1. When k = 1,
(2) represents perpendicular bisector of the segment
joining –2(1 – i) and 1 – i.
1 1
£ . Now
5. As |z| ≥ 5,
z 5
1
1 24
1
1
z- ≥ z = z - ≥ 5- =
z
z
z
5 5
The least value is attained when z = 5.
2p ˆ
2p
Ê
6. i Ë 1 - cos ¯ + sin
7
7
p
p
p
= 2 i sin 2 + 2 sin cos
7
7
7
p˘
Êpˆ È p
= 2 sin Ë ¯ Ícos + i sin ˙
7 Î
7
7˚
3p ˆ
Ê 3p
2 Ëcos
+ i sin ¯
4
4
17p
28
fi |(x + iy)2| |c + id| = |a + ib|
If |z| = 0, we get z = 0. If |z| = 1,
=
()
7. (x + iy)2 (c + id) = a + ib
|z| = 0 or |z| = 1
i ˆ
Ê 1
+
Also, i –1 = 2 ÁË ˜
2
2¯
È Ê 17p ˆ
Ê 17p ˆ ˘
+ i sin Ë
cos Ë
Í
¯
p Î
28
28 ¯ ˙˚
2 sin
7
2
=z
|z|2 = |z|
z - (1 - i )
=k
z + 2(1 - i )
()
Thus, arg (z) =
z is real and thus, Im(z) = 0
3. |z2| = | – |z||
fi
2015
È Ê 3p p ˆ
Ê 3p p ˆ ˘
cos Ë
- ¯ + i sin Ë
Í
p Î
4 7 ¯ ˙˚
4 7
2 sin
7
2
= tan [(w + w 2)p – p/3]
= tan (–p – p/3) = – tan(p + p/3)
= –tan (p/3) = – 3
11. Let z = –i t
where t > 0, then
i z = i(i t) = – t
\ arg(i z) + arg(z) = p – p/2 = p/2
12. The inequality a + ib < c + id is true if and only
if b = d = 0 and a > c.
13. Let z = x + iy, then z2 = x2 – y2 + 2ixy
\
Re(z2) = 0
fi
|Re(z)| = |Im(z)|
fi
x2 – y2 = 0
Êz ˆ
Êz ˆ
14. arg Á 1 ˜ + arg Á 2 ˜
Ë z4 ¯
Ë z3 ¯
Ê z 2ˆ
Êzz ˆ
= arg Á 1 2 ˜ + 2kp = arg Á 1 2 ˜ + 2kp = 2kp
Ë z4 z3 ¯
Ë z3 ¯
where k = 0 or 1
[ z1 = z3 = i, gives answer 2p
and z1 = z3 = 1, gives answer 0]
2.46
15.
Complete Mathematics—JEE Main
z-4 x-4 y
=
+
2i
2i
2
For
i
(– 1 + 2i)
2
Ê z - 4ˆ p
0 £ sin -1 Ë
£ , we must
2i ¯ 2
y
£1
2
0 £ y £ 2.
0£
x – 4 = 0,
fi
or z = (2 – i) –
x = 4,
-3i
fi z is purely imaginary.
z +a
17. Let z = bi, b Œ R, b π 0,
16. z =
Fig. 2.56
Then z2 = –b2 Œ R,
3
1
fi z = 1 - i or z = 3 - i
2
2
Therefore Im(z2) = 0
18. We have zk = wk where
Êpˆ
Êpˆ
w = cos Ë ¯ + i sin Ë ¯
10
10
Thus, z1 z2 z3 z4 = w . w2. w3. w4 = w10
Ê 10 p ˆ
Ê 10 p ˆ
= cos Ë
+ i sin Ë
¯
10
10 ¯
[by the De Movire’s Theorem]
= cos p + i sin p = –1.
\ equation whose roots are a19, b 7 is x 2 + x + 1
= 0.
25. Using w10 = w, w 23 = w2, we get
{
sin (w 10 + w 23 ) p -
z1 + z2 1 / z1 + 1 / z2
z +z
= 1 2 =z
=
1 + z1 z2 1 + 1 / z1 z2 1 + z1 z2
Thus, z is purely real.
20. Let z = x + iy, then z2 = x2 – y2 – 2ixy
Re(z2) = k2 fi
}
p
1
p
p
= sin Ê -p – ˆ = sin =
Ë
¯
4
4
4
2
26. (1 + w)2017 = (– w 2)2017 = – w 4034 = – w2
\ A + Bw = 1 + w
19. z =
\
a = w, b = w 2 so that a 19 = w and b 7 = w 2.
24. Let
27. Use
A = 1, B = 1.
R2 Æ R2 – R1 – R3
28. (1 + w – w 2)7 = (– w 2 – w 2)7 = – 27w14 = – 128w2
29. Use
x2 – y2 = k2
fi
C2 Æ C2 + 3i C3.
30. Let a = arg z < 0, then arg(– z) = p + a,
which represents a hyperbola.
Level 1
21. |1 – w| = |w – w 2| = |w2 – 1|
Alternatively plot the points on an argand diagram.
3
( )
22. x3 = p = p 3
1
1
1
Let
1
1
fi x = p 3 , p 3 w, p 3 w 2.
1
1
xa + yb + zg x + yw + w 2 z 1
=
= = w2
2
x b + yg + za xw + yw + z w
1
3
1
3
1
3
2
If a = p w, b = p , g = p w , then
xa + yb + zg
1
=
= w.
x b + yg + za w 2
23.
Fig. 2.57
a = p 3 , b = p 3 w, g = p 3 w 2.
z - (2 - i )
z - (2 - i ) 1
AM ± p i 2
= (± i )
=
fi
e
- 1 + 2i
1 + i - (2 - i ) MD
2
fi z = (2 – i) +
i
(– 1 + 2i)
2
31.
fi
|z1| = |z2| = |z3| = 1
z1 z1 = z2 z2 = z3 z3 = 1
\ |z1 + z2 + z3| =
1
1
1
1
1
1
+
+
+
+
=
=1
z1 z2 z3
z1 z2 z3
32. nth roots of unity are given by
Ê 2mp ˆ
Ê 2mp ˆ
2mp i/n
cos ÁË
for m = 0, 1, 2,
˜¯ + i sin ÁË
˜=e
n
n ¯
, n – 1.
2m p i / n
2m p i / n
Let z1 = e 1
and z2 = e 2
where 0 £ m1, m2 < n, m1 π m2.
As the join of z1 and z2 subtend a right angle at the
origin z1/z2 is purely imaginary we get
Complex Numbers 2.47
e2 m1p i / n
e2 m2p i / n
= il for some real l fi
e2(m1 - m2 )p i / n = il
È 2 ( m1 - m2 ) p ˘
È 2 ( m1 - m2 ) p ˘
fi cos Í
˙ + i sin Í
˙ = il
n
n
Î
Î
˚
˚
2 ( m1 - m2 ) p p
È 2 ( m1 - m2 ) p ˘
fi cos Í
=0 fi
=
˙
n
2
n
Î
˚
fi n = 4(m1 – m2)
Thus, n must be of the form 4k.
1- i 3
z1 - z3
z -z
= 1 3 =
2
z2 - z3
z2 - z3
33.
=
fi
Also,
fi
fi
fi
1
(1 + 3) = 1
4
|z1 – z3| = |z2 – z3|
z1 - z3
1- i 3
–1=
–1
2
z2 - z3
z1 - z2
-1 - i 3
=
z2 - z3
2
| z1 - z2 |
=
| z2 - z3 |
1
(1 + 3) = 1
4
|z1 – z2| = |z2 – z3|
z 3 = z fi |z3| = | z | fi |z3| = | z |
38.
fi
|z| = 0 or |z| = 1 fi z = 0 or z = 1/z.
\
z3 = z fi z 3 =
fi
z = 1,–1, i, –i.
Thus, solutions of z 3 = z are 0, 1, –1, i, –i
39. Let
z = x + iy so that
x = 5 + t,
0
0
2
D = 1 w –1
w –1
1 w2 – 1
w –1
2
= (w – 1) – (w 2 – 1) 2
= (w + w2 – 2) (w – 1 – w2 + 1)
= (– 3) (w – w 2) = 3w (w – 1)
35. |z – i| = |z + i| represents the real axis.
As z = i satisfies |z – i| < |z + i|, we get
|z – i| < |z + i|, represents Im (z) > 0
36. iz 3 + z2 – z + i = 0 fi iz 3 – i2z2 – (z – i) = 0
fi iz 2 (z – i) – (z – i) = 0
fi (iz 2 – 1) (z – i) = 0
fi z 2 = 1/i or z = i
In any case |z| = 1.
2(1 - cos q )
37. 2x = (x + iy) + (x – iy) =
(1 - cos q )2 + 4 sin 2 q
fi
1
x=
5 + 3 cosq
Thus, maximum value of x is 1/2. It is attained at
q = p.
25 – t 2
¤ x [bc + ca + ab + 2(a + b +c)x + 3x2]
= 2[abc + (bc + ca + ab)x + (a + b + c)x 2 + x3]
¤ x 3 – (bc + ca + ab)x – 2abc = 0.
If a is the third root of this equation, then
a + w + w2 = 0 fi a = 1.
41. z = x + iy, |z – i Re(z)| = |z – Im(z)|
fi | x + iy – ix|2 = |x + iy – y|2
fi x2 + (y – x)2 = (x – y)2 + y2
fi x = ± y.
42. We have
Hence, z1, z2 and z3 are the vertices of an equilateral
1
y=
fi (x – 5) 2 + y2 = 25,
This clearly passes through 2 + 4i
40. Note that w, w2 are roots of
1
1
1
2
+
+
=
a+x b+x c+x x
Thus, |z1 – z3| = |z2– z3| = |z2 – z1|
triangle.
34. Using 1 + w 2 = – w, w4 = w and applying
C2 Æ C2 – C1, C3 Æ C3 – C1 we get
1
fi z4 = 1
z
(r – w) (r – w2) = r2 – (w + w2 ) r + 1
= r 2 + r + 1 = (r + 1)2 – r
12
12
\
 ( r – w ) (r – w 2 )
=
r =1
 ÈÎ(r + 1)2 – r ˘˚
r =1
1
1
= (13) (13 + 1) (26 + 1) – 1 – (12 ) (13) = 740
6
2
43. Let ix (x ΠR) be root of z2 + az + b = 0, then
(1)
– x 2 + aix + b = 0
fi – x 2 – a ix + b = 0
Subtracting we get
(a + a ) ix + b – b = 0
fi
x= –
i (b – b )
b–b
=
a+a
i (a + a )
Putting this in (1), we get
(b – b )2
2
(a + a )
–
a (b – b )
+b=0
a+a
fi (b – b )2 – a(b – b ) (a + a ) + b(a + a )2 = 0
fi (b – b )2 – (a + a ) {ab – a b – ab – a b} = 0
fi (b – b )2 + (a + a ) (a b + a b) = 0
44. The circle |z – (– 1 + i)| = 2 completely lies
inside the circle |z| = 3.
45.
8iz 3 + 12z2 – 18z + 27i = 0
fi 8iz 3 – 12i2z2 – 18z + 27i = 0
fi 4iz2 (2z – 3i) – 9 (2z – 3i) = 0
Complete Mathematics—JEE Main
2.48
(4iz2 – 9) (2z – 3i) = 0
fi
9
3i
z2 =
or z =
4i
2
fi
In any case |z| = 3/2.
46. As – 1 lies on the circle |z – 4| = 5, the real number
|z + 1| is maximum when z is the other end point
of the diameter.
y
–1
4
9
52. As |w| = 1 fi |w |2 = 1 fi ww = 1 fi
Thus,
Now,
1
= w.
w
z = w + 1/w = w + w = 2 Re(w)
|Re(z)| = |2 Re(w)| = |2 Re(w)| £ 2|w | = 2
53. | z | = 1 fi z = cos q + i sin q for some q Π[0, 2p)
Now,
| z | = 1 fi |z|2 = 1 fi zz = 1.
z z
1
+ = z2 + 2
z z
z
2
= (cos q + i sin q) + (cos q – i sin q)2
= 2(cos2q + i2 sin2q) = 2 (cos 2q)
Thus,
x
Fig. 2.58
47.
cosq + 2
1
=
+ i sinq
3
x + iy
3
cosq + 2
sinq
x – iy
fi 2
=
+ i
3
3
x + y2
x
fi
2
x +y
2
=
cosq + 2
3
Fig. 2.59
9
=
Also, x + y =
2
2
(2 + cos q ) + sin q 5 + 4 cos q
2
9
2
\ 4x – (x2 + y2)
4 (cosq + 2 ) ˆ 2
= Ê
- 1¯ ( x + y 2 )
Ë
3
=
4 cos q + 5
9
◊
=3
3
5 + 4 cos q
48. Affix of G is
1
( z1 + z2 + z3 ) .
3
As origin is the mid-point of AG,
1 1
0 = ÈÍ ( z1 + z2 + z3 ) + z1 ˘˙
˚
2 Î3
fi 4z1 + z2 + z3 = 0.
az + bz2
1
49. z3 = – ( az1 + bz2 ) = 1
c
a+b
50.
z –1
z –1
= it where t ΠR. fi
= – it
z +1
z +1
z –1
z –1
fi
+
=0
z +1
z +1
fi (z – 1) ( z + 1) + ( z – 1) (z + 1) = 0
fi 2(z z – 1) = 0 fi |z| = 1.
51.
z
z
–
z
z
£ |arg (z) – arg( z )|
fi |z – z | £ |z| |arg (z) – arg( z )|
z z
1
+ = 1 fi |cos 2q| =
z z
2
fi cos 2q = ± 1/2
fi 2q = p/3, 2p/3, 4p/3, 5p/3, 7p/3, 8p/3, 10p/3, 11p/3
fi q = p/6, p/3, 2p/3. 5p/6, 7p/6, 4p/3, 5p/3, 11p/6
Hence, there are 8 values of z
\
54. |z1 + z2|2 + |z1 – z2|2 = 2|z1|2 + 2|z2|2
fi
3 + |z1 – z2|2 = 2(2) + 2(2)
fi |z1 – z2|2 = 5 fi |z1 – z2| = 5 .
55. z1 z2 = z2 z3 = z3 z1
fi
|z1| |z2| = |z2| |z3| = |z3| |z1| fi |z1| = |z2| = |z3| = r(say)
[ z1, z2, z3 π 0]
Thus,
z1 z1 = z2 z2 = z3 z3
Now,
z1 z2 = z2 z3 = z3 z1
fi
z
z1
z
= 2 = 3
z2
z3
z1
fi
z12 = z2z3,
z22 = z3 z1 , z32 = z1 z2
Hence, z12 + z22 + z32 = z2z3 + z3z1 + z1z2
fi z1, z2, z3 are vertices of an equilateral triangle.
56. |z1 + z2 + z3| ≥ 0
fi |z1|2 + |z2|2 + |z3|2 + 2Re ( z1 z2 + z2 z3 + z3 z1 ) ≥ 0
fi
Re ( z1 z2 + z2 z3 + z3 z1 ) ≥ –
3
2
Now, |z2 – z3|2 + |z3 – z1|2 + |z1 – z2|2
= 6 – 2Re ( z1 z2 + z2 z3 + z3 z1 )
Complex Numbers 2.49
\ A = (–1, 1)
61. |z| = 1 fi zz = 1. Therefore z = 1 – 2z
fi z = 1 – 2/z fi z2 – z + 2 = 0
£ 6 – (–3) = 9
57. z1 + z2 + z3 = 0 fi z1 + z2 + z3 = 0
1
1
1
+
+
=0
z1 z2 z3
fi
[
2
Now, 0 = (z1 + z2 + z3)
= z12
z22
+
+
z32
z1 z1 = 1 etc.]
+ 2 (z2z3 + z3z1 + z1z2)
fi
= z12 + z22 + z32
63. z = i (1 + 3 ) = - 1 + 3 i = 2 w
= 0+
58. Note that
z22
= z2
2
where w π 1 is a cube root of unity.
\
=1
1
= z1 etc. Thus,
z1
z = (z1 + z2 + z3) ( z1 + z2 + z3 )
= |z1 + z2 + z3|2 £ (|z1| + |z2| + |z3|)2 = 9
Then z4 = a = b4(–1)
p
pˆ
Ê
fi z = b ˱ cos ± i sin ¯
4
4
\
z2 - 1 z - 1
1- z
= z -1+
=
z +1 z +1
1+ z
1
zˆ
Ê
Re(w) = Re( z ) - 1 + Re
Ë1+ z¯
\
Re
z4 + 2z3 + 4z2 + 5
= (2w)4 + 2 (2w)3 + 4 (2w)2 + 5
= 16w4 + 16w3 + 16w2 + 5
= 16 (w + 1 + w2) + 5 = 16 (0) + 5 = 5
64. Let a = – b4 where b > 0.
The maximum value is obtained when z1 = z2 = z3 = 1.
z-zz
z - 1 ( z - 1)z
=
=
59. w =
[ |z| = 1]
z +1
z + z z 1 z +1
But
but then |z| π 1.
2iz1 + 5z2
2i + 5 ( z2 / z1 ) 2i + 5a
=
=1
=
2iz1 - 5z2
2i - 5 ( z2 / z1 ) 2i - 5a
Now,
z12 + 2 z22 + z32 = z12 + z22 + z32 + z22
Thus,
1± 7 i
,
2
62. Suppose z2/z1 = a, where a ΠR.
Ê1
1
1ˆ
0 = z12 + z22 + z32 + 2z1z2z3 Á +
+
Ë z1 z2 z3 ˜¯
fi
z=
z12 + z22 + z32 + z42
È Êp ˆ
Êp ˆ
Êp ˆ
Êp ˆ˘
= 2b2 Ícos Ë ¯ - i sin Ë ¯ + cos Ë ¯ + i sin Ë ¯˙
Î
2
2
2
2 ˚
= 0 = a + |a|
[ a < 0]
Ê1- zˆ 1 Ê1- z 1- z ˆ
=
+
Ë1+ z¯ 2 Ë1+ z 1+ z ¯
1 Ê 1 - z z - 1ˆ
=0
+
=
2 Ë 1 + z z + 1¯
Re(w) = cosq –1 = –2 sin2(q/2)
z+z
is a positive real number,
1 + zz
Ê z+z ˆ
arg
= 0.
Ë 1 + zz ¯
66. Using (cos a + i sin a) (cos b + i sin b)
65. As
= cos (a + b) + i sin (a + b),
2
60. Let w =
1+ z
1
. Im(w) = (w - w )
2iz
2
2
For |z| = 1,
w=
=
Thus, Im(w) =
= -
we get
cos (25q) = – 1, sin (25 q) = 0
fi 25q = (2k+ 1)p, k ΠI.
2
1+ z
1 + (1 / z )
=
-2iz
-2i (1 / z )
z2 + 1
= –w
-2iz
1(
w + w ) = -iw
2i
fi
k = 0, 1, 2, …, 12
67. log(1/7)|z – 2| > log(1/7)|z|
fi
1Ê
1ˆ
1
= - (z + z )
z+
Ë
¯
2
z
2
1
= - (2 cosq ) = – cos q
2
[ |z| = 1 fi z = cos q + i sin q]
As z π 1, q π 0 and as z π –1, q π p,
fi
(2 k + 1)p
£p
25
1 £ (2k +1) £ 25
Now, 0 <
|z – 2| < |z|
(1)
But |z – 2| = |z| represents perpendicular bisector
of the segment joining 0 and 2, that is, |z –2| = |z|
represents the line Re(z) =1. As 0 does not satisfy
(1), we get (1) represents Re(z) > 1.
68.
1 È 2z + 1 2 z + 1 ˘
=5
2i ÍÎ iz + 1 -i z + 1 ˙˚
2.50
Complete Mathematics—JEE Main
= (–1)n+2 wn+2 + 1
fi
(2z + 1) (i z –1) + (2 z + 1) (iz +1)
= 10 i (iz + 1) (i z – 1)
fi 4iz z + (–2 + i) z + (2 + i) z
= 10 i (–z z – iz + i z – 1)
fi 14iz z + (8 + i) z + (–8 + i) z – 10 i = 0
5
1
1
fi z z + (1 - 8i )z + (1 + 8i ) z - = 0
14
14
7
This represents a circle.
69. Take z1 and z2 as two real numbers.
70.
(1 + i )
(1 - i )
n
n-2
Ê1+ iˆ
=Ë
1- i¯
n-2
2
Ê -i + i ˆ
(1 + i )2 = ÁË
˜
1- i ¯
( 2i )
Thus, range of f is {0, 3}.
1
72. z +
= 2 cos q fi z2 – 2z cos q + 1 = 0
z
fi
z = cos q ± i sin q
Now, z2n – 2 zn cos (nq)
= zn [zn – 2 cos (nq)]
= zn [cos(nq) ± i sin(nq) – 2 cos(nq)]
= –zn z n = –1
73. Using wk + wk+1 + wk+2 = 0 " k ΠI,
60
Âw
we get
=0
Next,
’w
k
fi
t = w, w2 where w π 1 is a cube root of unity.
z
=w
z -1
|z| = |z – 1| |w | = |z –1|
Thus,
fi z lies on the perpendicular bisector of the
segment joining z = 0 and z = 1, that is, z lies on
Re(z) = 1/2. Let z = 1/2 + ai where a ΠR.
Note that a π 0 since z = 1/2 does not satisfy z2 +
z|z – 1| + |z –1|2 = 0
Putting z = 1/2 + ai in z2 + z|z –1| + |z –1|2 = 0
and equating imaginary parts of both sides, we get
a + a|z –1| = 0 fi |z – 1| = –1.
A contradiction.
Thus z2 + z |z – 1| – |z + 1|2 = 0 has no solution in C,
that is, statement-1 is true.
Statement-2 is false as aw, aw2 are solutions of
z2 + az + a2 = 0.
77. z = 2i + z
fi
fi
z – z = 2i
2i Im(z) = 2i
fi
Im(z) = 1
As |z| = 1, we get z = ± i
As arg(i) = p/2 and arg (–i) = –p/2,
m
\
= w = where
k =1
Statement-2 is false.
78. Note that
1
m = 1 + 2 + … + 30 = (30)(31) = 465
2
30
\
t = z/|z –1|
Thus, statement-1 is true.
k =1
30
76. Clearly z π 0, 1. We can write the given equation
as t2 + t + 1 = 0 where
fi
n-2
= i n –2 (2i) = 2i n –1 = –2i n +1
71. If n = 3k, k ΠI, then f(n) = 1 + 1 + 1 = 3
If n = 3k + 1, k ΠI, then f(n) = 1 + w + w2 = 0
If n = 3k + 2 k ΠI, then f(n) = 1 + w2 + w = 0
k
= 0 if n = 6 k + 1
’w k
8p ˆ
Ê8p ˆ
Ê
a = cos Ë ¯ + i sin Ëp - ¯
11
11
Ê8p ˆ
Ê8p ˆ
= cos Ë ¯ + i sin Ë ¯
11
11
= w465 = 1
k =1
74. x2 + x + 1 = (x – w) (x – w2)
Therefore, w, w2 are zeros of P(x)
\ a11 = cos(8p) + i sin(8p) = 1
[De Moivre’s Theorem]
0 = g(1) + w h(1) and 0 = g(1) + w2h(1)
Now,
10
Âak =
k= 0
fi g(1) = 0, h (1) = 0.
\
2
Therefore, –w, –w2 must be zeros of xn +1– xn+1.
Now,
(–w)
n+1
Statement-2 is true
2
75. x – x + 1 = (x + w) (x + w )
n
– (–w) + 1 = (–1)
n+1
n
w (w + 1) + 1
= (–1)n+1 wn (–w2) + 1
1 - a 11
=0
1-a
We have a =
aa 1
= a10 etc.
=
a
a
Thus,
Re (a + a2 + a3 + a4 + a5)
Complex Numbers 2.51
1
(a + a 2 + a 3 + a 4 + a 5 +a + a 2 + a 3 + a 4 + a 5 )
2
=
=
1 10 k 1
1
a = (-1) = Â
2 k =1
2
2
[Use statement-2]
\ Statement-1 is also correct and Statement-2 is
a correct explanation for it.
Statement-2 is true as it is formula for sun of a G.P.
79. zk = cos (k q) + i sin(k q)
15
f(q) =
 Im (z2 k -1 )
k =1
Ê 15
ˆ
= Im Á Â z 2 k -1 ˜
Ë k =1
¯
Ê z(1 - z 30 ˆ
= Im Á
Ë 1 - z 2 ˜¯
But z30 = cos (30 q) + i sin (30 q)
When q = a = p/60, z30 = 0 + i = i
Ê z (1 - z 30 ) ˆ
\ f(a) = Im Á
Ë 1 - z 2 ˜¯
= Im
Ê 1- i ˆ
Ë 1/ z - z¯
Ê 1- i ˆ
= Im Á
Ë - 2i sin a ˜¯
1
1
=
Im (1 + i ) =
2 sin a
2 sin a
80. Statement-2 is true. See theory of chapter 5 on
matrices.
È 1 a b˘
Let A = Í w 1 c ˙ ,
Í
˙
ÍÎw 2 w 1 ˙˚
w c
w 1
1 c
-a 2
+b 2
w 1
w w
w 1
2
= 1 – cw – a(w – cw )
= (1 – aw) (1– cw)
Note that |A| = 0 if a = w2 or c = w2
Thus, |A| π 0 if a = c = w and b = w or w2
|A| = 1
\
S contains exactly two distinct elements.
81. |z1 – z2| –
1 1
z1 z2
z2 - z1
z1 z2
= |z1 – z2| – |z2 – z1| = 0
= z1 - z2 -
\
Statement-1 is true.
Statement-2 is also true but a correct explanation
for truth of statement-1.
82. Statement-2 is a true statement.
Suppose f (z) = (z2 + 1) q (z) + r (z).
If r (z) π 0, then r (z) = az + b
where a, b ΠC.
We have
i = f (i) = (i2 + i) q(i) + ai + b
fi ai + b = i
Also, 1 + i = f (–i) = ((–i)2 + 1) q (–i) – ai + b
fi – ai + b = 1 + i
1
1
\ b = + i, a = i
2
2
1
Thus, az + b = (iz + 1) + i
2
83. Statement-2 is true, as z3 = a3 fi |z|3 = |a|3
fi |z| = |a|
Next,
(z + ab)3 = a3
fi z + ab = a, aw, aw2
Let z1 = a – ab, z2 = aw – ab
and z3 = aw2 – ab
We gave |z2 – z1| = |a| |w –1| =
|z3 – z2| =
3 a , |z1 – z3| =
3a,
3a
Thus, statement-1 is also true, but statement-2 is
not a correct explanation of statement-1.
84. Statement-2 is true. See Theory. Let a be a root of
z2 – z + p = 0 and suppose |a| < 1. We have
p = |p| = |a – a2| £ |a| + |a|2 < 2
A contradiction.
Thus, statement-1 is also true and statement-2 is a
correct explanation for it.
85. We know that |sin q | £ q
" q ≥ 0.
If q < 0, then |sin (–q)| £ – q fi |sin q | £ |q |
Thus, |sin q | £ |q | " q Œ R and statement-2 is true.
Now, |z1 – z2|2 = (r1 cos q – r2 cosj )2
+ (r1 sin q – r2 sinj )2
= r12 + r22 - 2r1r2 cos(q - j )
Êq - j ˆ
2
= (r1 - r2 ) + 4r1r2 sin 2 Ë
2 ¯
2
Êq - f ˆ
£ (r1 – r2)2 + 4(1) (1) Ë
2 ¯
Thus, Statement-1 is also true and Statement-2 is
a correct explanation for it.
2.52
Complete Mathematics—JEE Main
y
Level 2
2
86. As ABCD is a parallelogram,
mid point of AC = mid point of BD
-2
1
1
fi ( z1 + z3 ) = ( z2 + z4 )
2
2
fi z4 = z1 + z3 – z2
x
2
-2
87. |a + bw + cw2|2 = (a + bw + c w ) (a + b w + cw)
= a2 + b2 + c2 – bc – ca – ab
1
2
2
2
= ÈÎ(b - c ) + (c - a ) + (a - b ) ˘˚
2
As a, b, c are integers and at least two of them are
unequal, we get,
0
Fig. 2.60
This represents a square. See Fig. 2.60.
91. Let z = x + iy, then
x = 1 – t, y =
t2 + t + 2
(b – c)2 + (c – a)2 + (a – b) ≥ 2.
fi t = 1 – x and y2 = t2 + t + 2 = (t + 1/2)2 + 7/4
Thus, |a + bw + cw2 |2 ≥ 1 fi |a + bw + cw2| ≥ 1
fi y2 = (x – 3/2)2 + 7/4.
Least value 1 is attained when a = 2, b = 1, c = 1.
This represents a hyperbola.
88. We have AP = AB = AQ = 2
Thus, for the shaded region |z + 1| > 2
p
Ê
2 ˆ
=
Also, –BAQ = tan -1 Á
˜
4
Ë 2 - 1 + 1¯
and –BAP = – p / 4
Hence, for the shaded region
|z + 1| > 2 and |arg (z + 1)| < p / 4
w - wz
89. As
is purely real,
1- z
w - wz
w - wz
=
1- z
1- z
fi (1 – z ) (w – wz ) = (1 – z) ( w – wz )
fi (w – w ) (1 – z z ) = 0
As w π w , we get z z = 1
Thus, set of values of z is
{z: |z| = 1, z π 1}.
90. zw = |z|2 fi zw = z z fi w = z
Thus, |z – z | + |w + w | = 4
fi |z – z | + | z + z| = 4
fi |2iy| + |2x| = 4
fi |x| + |y| = 2
92. w + w =
=
=
az1 - bz2 az1 - bz2
+
az1 + bz2 az1 + bz2
(az1 - bz2 ) (az1 + bz2 ) + (az1 - bz2 ) (az1 + bz2 )
az1 + bz2
2
a2 z1 z1 - b2 z2 z2
= 0 [ a |z1| = b |z2|]
2
az1 + bz2
fi w lies on the imaginary axis.
93. |z| + |1 – z| + |z – 2|
≥ max {|z|, |(1 – z) + (z – 2)|, |z + (1 – z)| +
|z – 2|, |1 – z| + |z – (z – 2)|} = 2
The value 2 is attained when z = 1
94. See Theory.
95. |z2 – z3|2 + |z1|2 = |z2 – z3|2 + | –z2 – z3|2
fi |z2 – z3|2 + 1 = 2|(|z2|2 + |z3|2) = 4
fi |z2 – z3| = 3
Similarly, |z3 – z1| = |z1 – z2| =
3
Thus, z1, z2, z3 are vertices of an equilateral
triangle.
96. 1 + w + w2 + º + wn – 1
=
=
=
1 - wn
1 - cos p + i sin p
=
1- w
1 - cos (p / n) - i sin (p / n)
2
2
2 sin (p 2n) - 2 i sin (p 2n) cos (p 2n)
2
- 2 i sin(p 2 n) [cos (p 2 n) + i sin (p 2 n)]
Complex Numbers 2.53
101. |z – a2| + |z – 2a| = 3 will represent an ellipse if
cos (p 2n) - i sin (p 2n)
Êpˆ
= 1 + i cot Á ˜
Ë 2n ¯
- i sin (p 2n)
=
|a2 – 2a| < 3
97. |z1 + z2|2 = |z1 – z2|2
¤ –3 < a2 – 2a < 3
fi |z1|2 + |z2|2 + z1 z2 + z1 z2
¤ –2 < (a – 1)2 < 4
= |z1|2 + |z2|2 – z1 z2 – z1 z2
¤ (a – 1)2 < 4 ¤ –1 < a < 3 ¤ aŒ(0, 3)
z1 z2
+
=0
z1 z2
98. z1 lies on the circle |z| = 2 and z2 lies on the line
x+y= 4 2
fi 2(z1 z2 + z1 z2) = 0 fi
Distance of x + y = 4 2 from (0, 0) is 4
Thus, minimum distance between z1 and z2 is 2.
102. As AB = BC = CA, we get
2|z| = |1| = |1 – 2z|
1
1
1
and |z – | =
fi |z| =
2
2
2
fi z is the point of intersection of circles
|z| = 1/2 and |z – 1/2| = 1/2
1
1 ± 3i
fiz=
4
103. |z + 1| + |z – 3| £ 10 represents the ellipse with
focii at (–1, 0), and (3, 0) and length of major axis
10. Its centre is (1, 0), and its equation is
(
z2
z1
( x - 1)2
25
+
)
y2
=1
21
Any point on the ellipse is P (1 + 5 cosq,
21 sin q ).
Its distance from A (7, 0) is given by
Fig. 2.61
AP2 = (5cosq + 8)2 + 21 sin2q
Alternative Solution
= (2cos q + 20)2 – 202 + 85
Distance of z1 = 2(cosq + isinq) from x + y = 4 2,
is
fi 182 – 202 + 85 £ AP2 £ 222 – 202 + 85
2 cos q + 2 sin q - 4 2
fi 3 £ AP £ 13
2
=
104. |z – 1|2 + 2| z – w|2 = 3|z – w2|2
2 ( 2 2 – (cosq + sinq))
fi |z|2 + 1 – z – z + 2[|z|2 + 1 – z w – z w]
Maximum possible value of cosq + sinq is 2 .
= 3[|z|2 + 1 – z w 2 – z w2]
fi (3w – 2w2 – 1) z + (3w2 – 2w – 1) z = 0
99. Let z1 = r (cosq + isinq).
Equation of tanget to x2 + y2 = r2 at (rcosq, rsinq)
is x cosq + y sinq = r
Ê z + z ˆ Ê z1 + z1 ˆ Ê z - z ˆ Ê z1 - z1 ˆ
fiÁ
= r2
+
Ë 2 ˜¯ ÁË 2 ˜¯ ÁË 2i ˜¯ ÁË 2i ˜¯
fi z z1 + z z1 = 2r 2
which represents a straight line.
105. As R.H.S is real, L.H.S must be real.
Ê 1 + i ˆ Ê i - i2 ˆ
Also, Á
=
Ë 1 - i ˜¯ ÁË 1 - i ˜¯
n
= in is real when n is even.
1 + x2 1 Ê
1ˆ
= Á x + ˜ > 1 for x π 1.
Ë
x¯
2x
2
Thus, we get only possible value of x is 1.
As x > 0 and
z
z
fi
+ =2.
z1 z1
100. Using |z1| + |z2| ≥ |z1 – z2|, we get
|z – 2 + 3i| + |z – 1 + i| ≥ |–1 + 2i| =
fi 9 £ AP2 £ 169
5
\ RHS = 1, thus, least value of n is 4.
Complete Mathematics—JEE Main
2.54
02 + z21 + z22 = 0(z1) + 0(z2) + z1z2
Previous Years’ AIEEE/JEE Main Questions
1. Let |z| = |w| = r and Arg (w) = q,
so that Arg (z) = p – q. We have
z = r[ cos (p – q) + i sin (p – q)]
= r[– cos q + i sin q]
= – r ( cos q – i sin q) = – w
7. z + iw = 0 fi z + iw = 0
Now, arg (zw) = p
fi z – iw = 0
Ê z2 ˆ
arg Á ˜ = p
Ë i ¯
2
fi arg (z ) – arg (i) = p
fi
Y
2.
(z1+ z2)2 = 3z1z2 fi a2 = 3b.
fi
fi 2 arg (z) – p/2 = p
fi
O
2
3
X
4
8. z
1/3
arg (z) = 3p/4
= p + iq
fi x – iy = (p + iq)3
If z satisfies |z – 4| = |z – 2|, then z lies on
the perpendicular bisector of the segment joining
z = 2 and z = 4.
i.e., |z – 4| = |z – 2| fi Re(z) = 3.
As z = 0 does not satisfy |z – 4| < |z – 2|, we get
|z – 4| < |z – 2| represents the region Re(z) > 3.
3. Suppose |z – w| = r touches |z – z1| = a and |z – z2|
= b externally.
fi
= p3 + 3p2(iq) + 3p (iq)2 + (iq)3
fi
x = p3 – 3pq2 and – y = 3p2q – q3
-y
x
= 3 p2 - q2
= p 2 - 3q 2 and
q
p
x
y
fi + = -2( p 2 + q 2 )
p q
fi
\
Ê x yˆ
ÁË p + q ˜¯
( p 2 + q 2 ) = -2
9. |z2 – 1| = |z|2 + 1 can be written as |z2 +(–1)|
= |z2| + |–1|
Then |w – z1| = a + r, |w – z2| = b + r
¤
fi
fi
¤
z2
is a non-negative real number.
-1
z2 is a non-positive real number.
¤
z lies on the imaginary axis.
|w – z1| – |w – z2| = a – b
w lies on a hyperbola with foci at z1 and z2
Alternative solution
Let z = x + iy then |z2 – 1| = |z|2 + 1, we get
|(x2 – y2 – 1) + 2ixy| = x2 + y2 + 1
4. As 1 = – i2,
x
x
x
2
Ê 1 + i ˆ = Ê -i + i ˆ = Ê i(1 - i ) ˆ = i x
1 = ÁË
˜
Á
˜¯
ÁË 1 - i ˜¯
Ë 1- i ¯
1- i
fi x = 4n for some n ΠN.
5. | zw | = | z || w | = |z||w| = |zw| = 1
Arg ( zw) = arg (w) + arg ( z )
= arg (w) – arg (z) = – p/2
\ zw = | zw|[cos(– p/2) + i sin (–p/2)] = – i
fi
( x 2 - y 2 - 1)2 + 4 x 2 y 2 = x 2 + y 2 + 1
fi
(x2 – y2)2 + 1 – 2 (x2 – y2) + 4x2y2
= (x2 + y2 + 1)2
fi
(x2 + y2)2 + 1– 2(x2 – y2) = (x2 + y2)2 + 1
+ 2(x2 + y2)
fi
\ z lies on the imaginary axis.
10. (x –1)3 = – 8
3
fi
Ê - x - 1ˆ = 1
Ë
2 ¯
fi
-
6. z1 + z2 = – a, z1Z2 = b
As 0 , z1, z2 for an equilateral triangle,
– 4x2 = 0 fi x = 0
x -1
= 1, w , w 2
2
Complex Numbers 2.55
fi
x = – 1, 1 – 2w, 1– 2w2
As – 1 is an end point of a diameter of the circle,
maximum possible value of |z + 1| is 6 which is
attained when z = – 7
1
11. |w| = 1 fi |z| = z - i
3
1
fi z is equidistant from z = 0 and z = i. Thus, z
3
lies on the perpendicular bisector of the segment
joining z = 0 and z = i/3. Therefore, z lies on a
straight line.
16. z =
z1, z2 lie on a ray through the origin O and
same side of the origin
¤
arg (z1) = arg (z2)
2 kp
2 kp
+ i cos
11
11
2
2k p ˆ
k
p
= i Ê cos
- i sin
Ë
11
11 ¯
k
= iw
2p
2p
where
w = cos
- i sin
11
11
10
2k p
2k p ˆ
Thus,
S = Â Ê sin
+ i cos
Ë
11
11 ¯
k =1
13. sin
11
Êw -w ˆ
= iÂwk = iÁ
Ë 1 - w ˜¯
k =1
10
w11 = cos 2p – i sin 2p = 1
But
\
S=–i
14. z2 + z + 1 = 0
fi z = w or w2
where w is complex cube root of unity.
Let
z = w, so that
1
= w2
z
Thus,
2
2
2
2
Ê z + 1 ˆ + Ê z 2 + 1 ˆ + Ê z3 + 1 ˆ + + Ê z6 + 1 ˆ
˜
ÁË
ÁË
˜
ÁË
˜
ÁË
˜
z¯
z2 ¯
z3 ¯
z6 ¯
= (–1)2 + (–1)2 + (2)2+ (–1)2 + (–1)2 + (2)2
= 12.
15. |z + 4| £ 3 represents the interior and boundary of
the circle with centre at (–4, 0) and radius = 3. See
Figure.
fi
4
4
£ |z| –
£
|z|
|z|
|z|2 – 2|z| – 4 £ 0
fi
(|z| – 1)2 £ 5
17. |z| –
12. |z1 + z2| = |z1| + |z2|
¤
1
1
1
-1
=
=
fi z =
i -1
i - 1 -i - 1 i + 1
fi
z-
4
=2
z
|z| £ 5 + 1
18. z is equidistant from A(1 + 0i), B(–1 + 0i) and C(0
+ i). Thus, z is circumcentre of DABC, that is, there
is exaclty one such z.
(1 + w)7 = (– w2)7 = (– w2)6(– w2) = – w2
19.
=1+w
\ A = 1, B = 1
20. As a, b ΠR, the roots of z2 + a z + b = 0 are of
the form
1 + ia, 1 – ia, where a Œ R, a π 0.
Now, b = (1 + ia)(1 – ia) = 1 + a2 > 1
fi b Œ (1, •)
z2
is real, we get
z -1
z2
z2
=
z -1 z -1
¤ z 2 ( z - 1) = z 2 ( z - 1)
¤ zz ( z - z ) - ( z - z )( z + z ) = 0
¤ ( z - z )( zz - z - z ) = 0
¤ z - z = 0 or zz - z - z = 0
21. As
fi
z lies on the real axis
or z lies on a circle through the origin.
22. |z| = 1 fi zz = 1
1+ z ˆ
Ê 1+ z ˆ
arg ÊÁ
= arg Á
Ë 1 + z ˜¯
Ë 1 + 1/ z ˜¯
= arg (z) = q
23. Let z2/z1 = ik where k is a real number.
2 z1 + 3z2
2 / 3 + z2 / z1
2 / 3 + ik
=
=
=1
2 z1 - 3z2
2 / 3 - z2 / z1
2 / 3 - ik
1 + z2
1
. Im (w ) = (w - w )
2iz
2
1 + z 2 1 + (1 / z ) 2
For |z| = 1, w =
=
-2iz
-2i (1 / z )
24. Let w =
=
z2 +1
= -w
-2iz
2.56
Complete Mathematics—JEE Main
1
(w + w ) = -iw
2i
1
= - (z + z )
2
Thus, Im(w) =
1Ê
1ˆ
= - Áz + ˜
Ë
2
z¯
1
= - (2 cosq ) = - cosq
2
[
|z| = 1 fi z = cos q + isin q]
As z π 1, q π 0 and as z π –1, q π p
\
a Œ (–1, 1)
25. z = 1 - z
fi z + z =1
fi 2 Re(z) = 1 fi
Re (z) =
1
p
= cos
2
3
As |z| = 1, Re(z) = cos(p/3),
Thus, arg (z) = ± p/3
\ statement-1 is false and statement-2 is true .
26. Im(z) + 1 = 0 Im(z) = –1
Let z = a – i
Now, z + 2 | z + 1 | +i = 0
2
fi a - i + 2 (a + 1) + 1 + i = 0
fi a2 = 2 (a2 + 2a + 2)
\ a=–2
Thus, z = – 2 – i fi
27. z +
fi
(a + 2)2 = 0
|z| =
5
1
1 3
≥|z|- ≥
2
2 2
Minimum value 3/2 of |z + 1/2| is attained when
z = Р2, and 3/2 Π(1, 2).
Also, z π 1 for otherwise w = w
fi
Im(w) = 0
Êz ˆ
29. arg Ê z1 ˆ + arg Á 2 ˜
ÁË z ˜¯
Ë z3 ¯
4
2
= arg Ê z1z2 ˆ = arg Ê | z1 | ˆ = 0
ÁË | z |2 ˜¯
ÁË z z ˜¯
3
4 3
z
i
30. Let
= ik , where k ΠR.
z +i
fi
z – i = ikz – k
fi
z(1– ik) = – k + i
-k + i
1 - ik
2
k +1
Note that | z |2 =
=1
1+ k2
fi zz = 1 fi z = 1 z
1
Thus, z + = z + z , which is a real number.
z
Also, z + z = 0
fi
Im (z) = ± sin (p/3)
fi a2 + 4a + 4 = 0
\ |z| = 1
z=
fi
2Re(z) = 0
fi
fi
z = ai for some a ΠR.
Re(z) = 0
But in this case
z -i
is a real number
z +i
Therefore, z + z π 0 .
31. (1+ ai)2 = x + iy
fi
1 – a2 + 2ai = x + iy
fi
1 – a2 = x, 2a = y
fi
fi
1 - ( y / 2) 2 = x
4 – y2 = 4x
fi
y2 + 4x – 4 = 0
32. z1 - 2 z2 = 1
2 - z1z2
¤ | z1 - 2 z2 |2 = | 2 - z1z2 |2
¤ | z1 |2 + 4 | z2 |2 - 2 z1z2 - 2 z1z2
w - wz = k(1 – z)
28.
fi
w – k = z(w - k )
fi
|w – k| = | z || w - k |
= |z||w – k|
As Im (w) π 0, |w – k| π 0
= 4 + | z1 |2 | z2 |2 - 2 z1z2 - 2 z1z2
¤ |z1|2 |z2|2 – |z1|2 – 4|z2|2 + 4 = 0
¤ (|z1|2 – 4) (|z1|2 – 1) = 0
As |z2| π 1, |z1|2 – 4 = 0 fi |z1| = 2
Complex Numbers 2.57
fi z1 lies on a circle of radius 2.
=0
fi 2(2 – 6 sin2q) = 0 fi sin q = ±
33. |z – 1| = |z + i|
fi
z is equidistant from A(1) and B(–i)
fi
z lies on the line y = – x
1
3
Thus, a value of q for which z is purely imaginary
-1 Ê 1 ˆ
is sin Á ˜
Ë 3¯
Coordinates of A are (2, 1) that of B are (3, 1) and
C are (3, 3).
Coordinates of D are (3 + 2 2 cos (5p/4), 3 +
2 2 sin (5p/4))
= (3 – 2, 3 – 2) = (1, 1)
Thus, D is represented by 1 + i
y
As z = 1 satisfies the inequality
3 + 3i
C
|z – 1| £ |z + i|,
2÷2
we get |z – 1| £ |z + i| represents the region lying
above the line x + y = 0
Largest value of r is length of perpendicular from
(4, 1) to the line x + y = 0, that is,
| 4 + 1| 5
5
=
=
2.
1+1
2 2
34. Let z = r (cos q + isin q), so that
largest r =
D
= 1 + 3ai – 3a2 – ia3
As z3 is real 3a – a3 = 0
fi a( 3 – a) (a +
z = r [cos (5q) + isin (5q)]
5
fia=
5
fi
(Im(z)) = r cos q
and
Im (z5) = r5 cos(5q)
p
pˆ
Ê
3 i = 2 Á cos + i sin ˜
Ë
3
3¯
Thus, z = 1 +
We have S = 1 + z + z2 + … + z11
Im( z 5 ) cos(5q )
=
(Im( z ))5
cos5 q
But cos (5q) = 16 cos5 q – 20 cos3 q + 5 cos q
1 - z12
1- z
=
But z12 = 212 (cos (4p) + i sin (4p))
cos(5q )
= 16 – 20 sec2 q + 5 sec4 q
cos5 q
= 5(sec4 q – 4sec2 q + 4) – 4
= 212(1 + i0) = 212
and 1 – z = –
3i
12
= 5(sec2 q – 2)2 – 4 ≥ – 4
Thus, S =
Thus, least value of
cos(5q )
is – 4 which is attained when q = p/4.
cos5 q
2 + 3i sin q
is purely imaginary, Re(z) = 0
35. As z =
1 - 2i sin q
that is, z + z = 0
2 + 3i sin q 2 - 3i sin q
fi
+
=0
1 - 2i sin q 1 + 2i sin q
fi (2 + 3isinq)(1 + 2isinq) + (2 – 3isinq)(1 – 2isinq)
3) = 0
3
Thus,
fi
x
37. z3 = (1 + ai)3
5
5
B
3+i
O
fi As z is non-real complex number, Im (z) π 0
5
2+i
=
1- 2
- 3i
4095
3
i = 1365 3 i
Previous Years’ B-Architecture Entrance
Examination Questions
1.
5 + i sin q
is a real number
5 - 3i sin q
¤
5 + i sin q
5 - i sin q
=
5 - 3i sin q 5 + 3i sin q
2.58
Complete Mathematics—JEE Main
¤
(5 + i sin q)(5 + 3 isin q)
= (5 – 3 isin q)(5 – i sin q)
¤
25 + 20 isinq – 3sin2q
= 25 – 20 isinq – 3sin2q
¤ 40 isin q = 0 ¤ sin q = 0
This is possible when q = – p.
TIP Amongst the choices given, – p is the only
value which makes imaginary part of the numerator
and denominator 0.
2. |z| = 2
Let w = 3z + 2 + i, then z = (w – (2 + i))/3
fi
|z| = 2
6. Area of triangle is
z
1
D = | iz
4
z + iz
z
1
-iz 1 |
z - iz 1
Using R3 Æ R3 – R1 – R2, we get
z
z
1
1
D = | iz -iz
1|
4
0
0 -1
1
1
| -izz - izz |= | z |2
4
2
7. As |i – (1 + 2i)| = 2 < | 4 – 2 | = 2,
=
C1 lies inside C2
|w – (2 + i)| = 6
fi w lies on a circle with centre at (2 + i) and
radius 6.
3. Let z = 2(cos q + i sin q). Now
z+
2
= 2(cos q + isin q) + (cos q – isin q)
z
= 3 cos q + isin q
fi
z+
22
= 9 cos2q + sin2q
z
= 1 + 4(1 + cos 2q) £ 9
2
is 3 which
\ maximum possible value of z +
z
is attained when z = 1
1
3
then w3 = 1
4. Let w = – + i
2
2
(
)
8. z = i i + 2 = –1 +
fi
2
(z + 1) = –2
2i
fi z2 + 2z + 3 = 0
We now divide
z4 + 4 z3 + 6 z
2
+ 4 z by z2 + 2z + 3.
z = 4 + 3w127 + 5w124
= 4 + 3(w3)42w + 5(w3)41w
= 4 + 3w + 5w = 4 + 8w
= 4 - 4 + 4 3i = 4 3i
5. Principal argument
is least at point P.
sin(p / 2 - q ) =
AP
OA
fi
cos q = 1/5
fi
sin q = 24 / 5
Now, z = 24(cosq + i sin q )
fi
Re(z) =
24 2
=
6
5
5
\
z4 + 4z3 + 6z2 + 4z
= (z2 + 2z + 3) (z2 + 2z – 1) + 3
=0+3=3
9. |z| = 1, arg(z) = q
fi z = cos q + isin q
Now, w =
z (1 - z ) z - zz
=
z (1 + z ) z + zz
Complex Numbers 2.59
=
w=
Now,
z -1
z +1
2
10.
2 + i = z1 + z2 + z3
2
fi 3 = |z1|2 + |z2|2 + |z3|2 + 2Re ( z1 z2 + z2 z3 + z3 z1 )
z -1
z +1
fi
2Re(w) = w + w =
z -1 z -1
+
z +1 z +1
z2 -1 + z 2 -1
=
( z + 1)( z + 1)
3 = 1 + 1 + 1 + 2 Re ( z1 z2 + z2 z3 + z3 z1 )
fi Re ( z1 z2 + z2 z3 + z3 z1 ) = 0
11. z (iz1 – 1) = z1 + 1
fi z1(iz – 1) = 1 + z
2 cos 2q - 2
=
zz + z + z + 1
fi z1 =
2(cos 2q - 1)
=
2(cosq + 1)
fi
1+ z
iz - 1
1+ z
= |z1| < 1
iz - 1
fi |1 + z|2 < |1 – iz|2
2
=
-2 sin (q )
qˆ
= - 4 sin 2 ÊÁ ˜
2
Ë 2¯
2 cos (q / 2)
fi 1 + z + z + zz < 1 – iz + iz + zz
fi 2 Re(z) < – i(2i Im(z))
fi
Êqˆ
Re(w) = -2 sin 2 Á ˜
Ë 2¯
fi Re(z) – Im(z) < 0
Quadratic Equations 3.1
CHAPTER THREE
5. If equation (1) is satisfied by more than two distinct
complex numbers, then equation (1) becomes an
identity, that is a = b = c = 0.
QUADRATIC EQUATIONS
An equation of the form
ax 2 + bx + c = 0
(1)
where a π 0, a, b, c Œ C, the set of complex numbers, is
called a quadratic equation.
A root of the quadratic equation (1) is a complex number
a such that
aa2 + ba + c = 0
The quantity D = b2 – 4ac is known as the discriminant of
the equation (1).
The roots of equation (1) are given by the quadratic formula
x=
-b ± D
2a
If a and b are the roots of the quadratic equation (1), then
b
c
a+b=–
and a b =
a
a
Note that a quadratic equation whose roots are a and b is
given by
(x – a) (x – b) = 0
QUADRATIC EXPRESSION AND ITS
GRAPH
NATURE OF ROOTS
1. If a, b, c Œ R and a π 0. Then the followings hold
good:
(a) The equation (1) has real and distinct roots if and
only if D > 0.
(b) The equation (1) has real and equal roots if and
only if D = 0.
(c) The equation (1) has complex roots with nonzero imaginary parts if and only if D < 0.
(d) p + iq (p, q Œ R, q π 0) is a root of equation (1)
if and only if p – iq is a root of equation (1).
2. If a, b, c ΠQ and D is a perfect square of a rational
number, then equation (1) has rational roots.
3. If a, b, c ΠQ and p +
RELATION BETWEEN ROOTS AND
COEFFICIENTS
q (p, q ΠQ) is an irrational
root of equation (1) then p – q is also a root of
equation (1).
4. If a = 1, b, c ΠI, the set of integers, and the roots
of equation (1) are rational numbers, then these
roots must be integers.
Let
We have
f (x) = ax2 + bx + c, where a, b, c Œ R and a π 0.
b
È
f (x) = a Í x 2 + x +
a
Î
c˘
a ˙˚
È
b
b2
c
b2 ˘
= a Í x2 + x + 2 + - 2 ˙
a
a 4a ˚
4a
Î
ÈÊ
b ˆ 2 4 ac - b2 ˘
= a ÍÁ x + ˜ +
˙
2a ¯
4a2 ˚
ÎË
(2)
When is a Quadratic Expression always positive
Or always negative?
It follows from (2) that f (x) > 0 (< 0) "x ΠR if and only if
a > 0 (< 0) and D = b2 – 4ac < 0. See Fig. 3.1 and (Fig. 3.2).
Also, it follows from (2) that f (x) ≥ 0 (£ 0) "x Œ R if and
only if a > 0 (< 0) and D = b2 – 4ac = 0. In this case f (x) > 0
(< 0) for each x Œ R, x π – b/2a. Also, in this case the graph of
y = f (x) will touch the x-axis at x = – b/2a.
3.2
Complete Mathematics—JEE Main
If D = b2 – 4ac > 0 and a < 0 then
Ï< 0 for x < b or x > a
Ô
f (x) Ì> 0 for b < x < a
Ô= 0 for x = a , b
Ó
See Fig. 3.4
Fig. 3.1
Fig. 3.4
Note that if a > 0, then f (x) attains the least value at x =
– b/2a. This least value is given by
2
Ê b ˆ 4 ac - b
f Á- ˜ =
Ë 2a ¯
4a
Fig. 3.2
If a < 0, then f (x) attains the a greatest value at x = – b/2a.
This greatest value is given by
Sign of a Quadratic Expression
If D = b2 – 4ac > 0, then (2) can be written as
2
È
2
Ê b2 - 4ac ˆ ˘
b
Ê
ˆ
f (x) = a ÍÁ x + ˜ - Á
˜ ˙
ÍË
2a ¯
2a
Ë
¯ ˙˚
Î
ÈÊ
b + b2 - 4ac ˆ Ê
b - b2 - 4ac ˆ ˘
= a ÍÁ x +
˜ Áx +
˜˙
2a
2a
ÍÎË
¯Ë
¯ ˙˚
= a (x – a) (x – b)
where a =
If
- b - b2 - 4ac
2a
and b =
D = b2 – 4ac > 0 and a > 0 then
Ï> 0 for x < a or x > b
Ô
f (x) Ì< 0 for a < x < b
Ô= 0 for x = a , b
Ó
- b + b2 - 4ac
2a
2
Ê b ˆ 4 ac - b
f Á- ˜ =
Ë 2a ¯
4a
POSITION OF ROOTS OF A
QUADRATIC EQUATION
Let f (x) = ax2 + bx + c, where a, b, c ΠR be a quadratic
expression and let k be a real number.
Conditions for Both the Roots to be more than a Real
Number k.
If a > 0 and b2 – 4ac > 0 then the parabola y = ax2 + bx +
c opens upwards and intersect the x-axis in a and b where
- b ± b2 - 4ac
2a
In this case both the roots a and b will be more than k if
k lies to left of both a and b. See Fig. 3.5
a, b =
See Fig. 3.3
Y
Fig. 3.3
Fig. 3.5
Quadratic Equations 3.3
From the Fig. 3.5, we note that both the roots are more
than k if and only if
-b
(i) D > 0
(ii) k <
(iii) f (k) > 0
2a
In case a < 0, both the roots will be more than k (see Fig.
3.6) if and only if
-b
(i) D > 0
(ii) k <
(iii) f (k) < 0
2a
(i) D > 0
(ii) af (k) < 0
Fig. 3.7
Fig. 3.6
Combining the above two sets, we get both the roots of ax2
+ bx + c = 0 are more than a real number k if only if
-b
(iii) af (k) > 0
(i) D > 0
(ii) k <
2a
Conditions for Both the Roots to be less than a Real
Number k
Both the roots of ax2 + bx + c = 0 are less than a real number
k if and only if
-b
(iii) af (k) > 0
(i) D > 0
(ii) k >
2a
Note
Fig. 3.8
CONDITIONS FOR EXACTLY ONE
ROOT OF A QUADRATIC EQUATION
TO LIE IN THE INTERVAL (k1, k2)
WHERE k1 < k2
If a > 0, then exactly one root of f (x) = ax2 + bx + c = 0 lies
in the interval (k1, k2) if and only if f(k1) > 0 and f (k2) < 0.
Also, exactly one root lies in the interval (k1, k2) if and only
if f (k1) < 0 and f (k2) > 0. See Fig. 3.9. Thus, if a > 0, exactly
one root of f (x) = ax2 + bx + c = 0 lies in the interval (k1, k2)
if and only if f (k1) f (k2) < 0.
1. Both the roots of ax2 + bx + c = 0 are positive if and only
if
b
c
D ≥ 0, - > 0 and > 0
a
a
2. Both the roots of ax2 + bx + c = 0 are negative if and only if
c
b
D ≥ 0, - < 0 and > 0
a
a
Fig. 3.9
CONDITIONS FOR A NUMBER k TO
LIE BETWEEN THE ROOTS OF A
QUADRATIC EQUATION
Similarly, if a < 0, exactly one of the roots of f (x) = ax2 +
bx + c = 0 lies in the interval (k1, k2) if f (k1) f (k2) < 0.
The real number k lies between the roots of the quadratic
equation f (x) = ax2 + bx + c = 0 if and only if a and f(k) are
of opposite signs, that is, if and only if
(i) a > 0
(ii) D > 0
(i) a < 0
(ii) D > 0
(iii) f (k) < 0 [See Fig. 3.7]
or
(iii) f (k) > 0 [See Fig. 3.8]
Combining, we may say k lies between the roots of f (x)
= ax2 + bx + c = 0 if and only if
Fig. 3.10
Hence, if f (k1) f (k2) < 0, then exactly one root of
f (x) = 0 lies in the interval (k1, k2).
3.4
Complete Mathematics—JEE Main
CONDITIONS FOR BOTH THE ROOTS
OF A QUADRATIC EQUATION TO LIE
IN THE INTERVAL (k1, k2) WHERE
k1 < k2.
Illustration
1
1 is a root of (b – c) x2 + (c – a) x + a – b = 0
as (b – c) + (c – a) + (a – b) = 0.
If a > 0, both the roots of f(x) = ax2 + bx + c = 0 lie in the
interval (k1, k2) if and only if
b
< k2
(i) D > 0
(ii) k1 < 2a
(iii) f(k1) > 0 and f(k2) > 0
See Fig. 3.11
Finding Range of a Rational Function
Let f(x) = ax2 + bx + c, g(x) = px2 + qx + r
where a, b, c, p, q, r Œ R and one of a, p π 0.
To find range of
f ( x)
, x ΠR, put
R(x) =
g( x )
fi
ax 2 + bx + c
,
px 2 + qx + r
(a – py)x2 + (b – qy)x + (c – r y) = 0.
Since x is real,
(b – qy)2 – 4(a – py) (c – r y) ≥ 0
We use this inequality to obtain range of R(x).
y=
Fig. 3.11
Illustration
In case a < 0, the conditions read as
b
< k2
(i) D > 0
(ii) k1 < 2a
(iii) f(k1) < 0 and f(k2) < 0
Find range of
x2 - 2 x + 5
y= 2
x + 2x + 7
See Fig. 3.12
fi
fi
fi
fi
fi
fi
Fig. 3.12
¤ D ≥ 0, -
(x2 + 2x + 7)y = x2 – 2x + 5
( y – 1)x2 + 2x(y + 1) + 7y – 5 = 0
As x is real,
4(y + 1)2 – 4(y –1) (7y – 5) ≥ 0
(y2 + 2y + 1) – (7y2 – 12y + 5) ≥ 0
3y2 – 7y + 2 £ 0
(3y – 1) (y – 2) £ 0
1/3 £ y £ 2
Some Useful Tips
Some Useful Tips
Let quadratic equation be
f(x) = ax2 + bx + c = 0
where a, b, c Œ R, a π 0.
Let roots of (1) be a, b, and D = b2 – 4ac.
1. Both the roots are positive
2
(1)
c
b
= a + b > 0,
= ab > 0
a
2a
ax 2 + bx + c
Suppose range of
px 2 + qx + r
real values, then range of
Illustration
a(h ( x ))2 + b (h( x )) + c
p(h ( x ))2 + q(h ( x )) + r
Range of
tan 2 x - 2 tan x + 5
2
tan x + 2 tan x + 7
3. Both the roots lie in the interval (p, q).
b
¤ D ≥ 0, p < <q, af( p) > 0, af(q) > 0
2a
4. If one of a, b is real, then other must be real.
5. If sum of the coefficients of a quadratic equation is 0,
then 1 is a root of the quadratic equation.
is also S.
3
2. Both the roots are negative
c
b
= a + b < 0, = ab > 0
¤ D ≥ 0, a
2a
is the set S and h(x) takes all
However, range of
is
1
£ y £ 2.
3
sin 2 x - 2 sin x + 5
sin 2 x + 2 sin x + 7
is not
[1/3, 2] as – 1 £ sin x < 1 will curtail its range. For
instance try y = 2.
Quadratic Equations 3.5
CONDITIONS FOR A QUADRATIC
EQUATION TO HAVE A
REPEATED ROOT
The quadratic equation f (x) = ax2 + bx + c = 0, a π 0 has a
as a repeated if and only if f (a) = 0 and f ¢(a) = 0. In this
case f(x) = a(x – a)2. In fact a = – b/2a. See Fig. 3.13 and
Fig. 3.14.
Fig. 3.15
TIP
Fig. 3.13
How to Obtain the Common Root?
Make coefficients of x2 in both the equations same and
subtract one equation from the other to obtain a linear
equation in x. Solve it for x to obtain the common root.
CONDITION FOR TWO QUADRATIC
EQUATIONS TO HAVE THE SAME
ROOTS
Fig. 3.14
CONDITION FOR TWO QUADRATIC
EQUATIONS TO HAVE A COMMON
ROOT
Suppose that the quadratic equations ax2 + bx + c = 0 and
a¢ x2 + b¢ x + c¢ = 0 (where a, a ¢ π 0 and ab¢ – a¢b π 0) have
a common root. Let this common root be a. Then
aa2 + ba + c = 0 and a¢a2 + b¢a + c¢ = 0
Solving the above equations, we get
a2
a
1
=
=
bc ¢ - b ¢c a ¢ c - ac ¢ ab ¢ - a ¢b
fi
bc ¢ - b ¢c
a ¢ c - ac ¢
and a =
a2 =
ab ¢ - a ¢b
a b ¢ - a ¢b
Eliminating a we get
(a ¢ c - ac ¢ )2
(ab ¢ - a ¢b)
fi
2
=
bc ¢ - b ¢c
a b ¢ - a ¢b
(a ¢c - ac ¢ )2 = (bc ¢ - b ¢c) (ab ¢ - a ¢b)
This is the required condition for two quadratic equation
to have a common root. See Fig. 3.15
Two quadratic equations ax2 + bx + c = 0. and a¢x2 + b¢x +
c¢ = 0 have the same roots if and only if
a
b c
=
=
a¢
b¢ c¢
EQUATIONS OF HIGHER DEGREE
The equation
f (x) = a0 xn + a1xn – 1 + a2xn– 2 + … + an–1 x + an = 0
where a0, a1, …, an – 1, an Œ C, the set of complex numbers,
and a0 π 0, is said to be an equation of degree n. An equation
of degree n has exactly n roots. Let a1, a2, …, an Œ C be
the n roots of (1). Then
f (x) = a0(x – a1) (x – a2) … (x – an)
Also
a1 + a2 + … + an = -
and
a1a2 … an = (–1)n
a1
a0
an
.
a0
CUBIC AND BIQUADRATIC
EQUATIONS
If a, b, g are the roots of ax3 + bx2 + cx + d = 0, then
b
c
d
a + b + g = – , bg + g a + ab =
and a b g = – .
a
a
a
3.6
Complete Mathematics—JEE Main
Also, if a, b, g, d are the roots of the equation ax4 + bx3 +
cx2 + dx + e = 0, then
–b
c
, (a + b ) (g + d ) + ab + g d =
a+b+g+d=
a
a
–d
e
ab (g + d) + gd ( a + b) =
, abgd = .
a
a
TRANSFORMATION OF EQUATIONS
We now list some of the rules to form an equation whose
roots are given in terms of the roots of another equation.
Let given equation be
(1)
a0 xn + a1 xn–1 + … + an–1 x + an = 0
Rule 1: To form an equation whose roots are k(π 0) times
x
roots of the equations in (1), replace x by
in (1).
k
Rule 2: To form an equation whose roots are the negatives
of the roots in equation (1), replace x by – x in (1).
Alternatively, change the sign of the coefficients of
xn–1, xn–3, xn–5, … etc. in (1).
Rule 3: To form an equation whose roots are k more than
the roots of equation in (1), replace x by x – k in (1).
Rule 4: To form an equation whose roots are reciprocals of
1
in (1) and
the roots in equation (1), replace x by
x
n
then multiply both the sides by x .
Rule 5: To form an equation whose roots are square of the
roots of the equation in (1) proceed as follows:
x in (1)
Step 1
Replace x by
Step 2
Step 3
Collect all the terms involving x on one side.
Square both the sides and simplify.
For instance, to form an equation whose roots are squares of
the roots of x3 + 2x2– x + 2 = 0, replace x by x to obtain
x x + 2x – x + 2 = 0
fi
x (x – 1) = –2(x + 1)
Squaring we get
x(x – 1)2 = 4(x + 1)2
or
x3 – 6x2 – 7x – 4 = 0
Rule 6: To form an equation whose roots are cubes of the
roots of the equation in (1) proceed as follows:
Step 1 Replace x by x1/3.
Step 2 Collect all the terms involving x1/3 and x2/3 on one
side.
Step 3 Cube both the sides and simplify.
DESCARTES RULE OF SIGNS FOR THE
ROOTS OF A POLYNOMIAL
Rule 1: The maximum number of positive real roots of a
polynomial equation
f(x) = a0 xn + a1 xn–1 + a2 xn–2 + … + an–1 x + an = 0
is the number of changes of the signs of coefficients
from positive to negative and negative to positive.
For instance, in the equation x3 + 3x2 + 7x – 11 = 0
the signs of coefficients are
+++–
As there is just one change of sign, the number of
positive roots of x3 + 3x2 + 7x – 11 = 0 is at most 1.
Rule 2: The maximum number of negative roots of the polynomial equation f(x) = 0 is the number of changes
from positive to negative and negative to positive in
the signs of coefficients of the equation f(– x) = 0.
SOME HINTS FOR SOLVING
POLYNOMIAL EQUATIONS
1. To solve an equation of the form
(x – a)4 + (x – b)4 = A
a+b
put
y =x–
2
In general to solve an equation of the form
(x – a)2n + (x – b)2n = A
a+b
where n is a positive integer, we put y = x –
.
2
2. To solve an equation of the form
(1)
a0 f(x)2n + a1(f(x))n + a2 = 0
we put (f(x))n = y and solve a0 y2 + a1y + a2 = 0
to obtain its roots y1 and y2.
Finally, to obtain solutions of (1) we solve,
(f(x))n = y1
and
(f(x))n = y2
3. An equation of the form
(ax2 + bx + c1) (ax2 + bx + c2) … (ax2 + bx + cn) = A
where c1, c2, …, cn , A Œ R, can be solved by putting ax2 + bx = y.
4. An equation of the form
(x – a) (x – b) (x – c) (x – d) = Ax2
where ab = cd, can be reduced to a product of two
ab
.
quadratic polynomials by putting y = x +
2
5. An equation of the form
(x – a) (x – b) (x – c) (x – d) = A
where a < b < c < d, b – a = d – c
can be solved by change of variable
y=
( x - a ) + ( x - b) + ( x - c) + ( x - d )
4
1
= x - (a + b + c + d)
4
6. A polynomial f (x, y) is said to be symmetric if
f (x, y) = f (y, x) ⁄ x, y.
A symmetric polynomials can be represented as a
function of x + y and xy.
Quadratic Equations 3.7
Equations Reducible to Quadratic
4
2
1. To solve an equation of the type ax + bx + c = 0,
put
x2 = y.
2. To solve an equation of the type
a (p(x))2 + bp(x) + c = 0
(where p(x) is an expression in x) put p(x) = y.
3. To solve an equation of the type
b
+c=0
ap(x) +
p ( x)
where p(x) is an expression in x, put p(x) = y.
This reduces the equation to
ay2 + cy + b = 0.
4. To solve an equation of the form
1ˆ
1ˆ
Ê
Ê
a Á x2 + 2 ˜ + b Á x + ˜ + c = 0
Ë
Ë
¯
x¯
x
1
put
x+
=y
x
and to solve
1ˆ
1ˆ
Ê
Ê
a Á x2 + 2 ˜ + b Á x - ˜ + c = 0
Ë
¯
Ë
x¯
x
1
put
x–
=y
x
5. To solve reciprocal equation of the type
ax4 + bx3 + cx2 + bx + a = 0, a π 0
we divide the equation by x2, to obtain
1ˆ
1ˆ
Ê
Ê
a Á x2 + 2 ˜ + b Á x + ˜ + c = 0
Ë
Ë
¯
x¯
x
1
= y.
and then put
x+
x
6. To solve equation of the type
(x + a) (x + b) (x + c) (x + d) + k = 0
where a + b = c + d, put x2 + (a + b)x = y
7. To solve equation of the type
ax + b = cx + d
or
ax 2 + bx + c = dx + e
square both the sides and solve for x.
8. To solve equation of the type
ax + b ± cx + d = e
proceed as follows.
Step 1 Transfer one of the radical to the other side and
square both the sides.
Step 2 Keep the expression with radical sign on one side
and transfer the remaining expression on the other side.
Step 3 Now solve as in 7 above.
USE OF CONTINUITY AND
DIFFERENTIABILITY TO FIND ROOTS
OF A POLYNOMIAL EQUATION
Let f (x) = a0 xn + a1 xn – 1 + a2xn – 2 + … + an – 1 x + an, then f
is continuous on R.
Since f is continuous on R, we may use the intermediate
value theorem to find whether or not f has a real root in an
interval (a, b).
If there exist a and b such that a < b and f (a) f (b) < 0,
then there exists at least one c Π(a, b) such that f (c) = 0.
Also, if lim f ( x ) and f (a) are of opposite signs, then at
xÆ - •
least one root of f (x) = 0 lies in the interval (– •, a). Also,
if f(a) and lim f ( x ) are of opposite signs, then at least one
xÆ •
root of f(x) = 0 lies in the interval (a, •).
Result 1 If f (x) = 0 has a root a of multiplicity r (where
r > 1), then we can write
f (x) = (x – a)r g(x)
where g(a) π 0.
Also, f ¢(x) = 0 has a as a root with multiplicity r – 1.
Result 2 If f (x) = 0 has n real roots, then f ¢(x) = 0 has
(n – 1) real roots.
It follows immediately using Result 1 and Rolle’s
Theorem.
Result 3 If f(x) = 0 has n distinct real roots, we can write
f(x) = a0 (x – a1) (x – a2) … (x – an)
where a1, a2, … an are n distinct roots of f(x) = 0.
We also have
n
f ¢ (x)
1
= Â
f (x)
k = 1 x - ak
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Example 1: Solution set of
3 - x = Рx2 Рx Р1, x ΠR is
(a) (– 1, •)
(b) (0, 1)
(c) (–1, 3/4)
(d) f
Ans. (d)
Solution: Note that LHS =
Ê
and RHS = – Ëx -
2
3 - x ≥ 0 ⁄ x £ 3.
1ˆ
5
- < 0 ⁄ x.
¯
2
4
3.8
Complete Mathematics—JEE Main
Thus, the given equation has no solution.
Example 2: Number of solutions of
x–
5
5
=2–
x-2
x-2
(1)
is
(a) 0
(c) 2
Ans. (a)
(b) 1
(d) infinite
fi
Example 3: The number of real solutions of x2 + 5|x| +
4 = 0 is
(a) 0
(b) 1
(c) 2
(d) 4
Ans. (a)
Solution: If x Œ R, then x2 + 5|x| + 4 ≥ 4.
Thus, the given equation has no solution.
Example 4: If
1
is a root of ax2 + bx + c = 0, (a, b,
3 - 4i
c Œ R, a π 0), then
(a) b + 6c = 0
(c) a + 25c = 0
Ans. (a)
(b) b = 6c
(d) b2 = 6c
Solution: TIP: As a, b, c ΠR, the other root must
Thus,
1
.
3 + 4i
1
1
b
+
=- ,
3 - 4i 3 + 4i
a
1
1
c
◊
=
3 - 4i 3 + 4 i a
b
6
1 c
=
= - and
a
25
25 a
b
Ê cˆ
Now, 6 Ë ¯ = - fi b + 6c = 0.
a
a
Also, a – 25c = 0
fi
Example 5: If 1 – p is a root of the quadratic equation
x + px + 1 – p = 0, then its roots are
(a) 0, 1
(b) –1, 1
(c) 0, – 1
(d) –2, 1
Ans. (c)
2
2
1
p (1 – p) < 0
3
p(1 – p) < 0 fi 0 < p < 1.
Solution: We must have
Solution: The equation (1) is defined when x π 2. But
when x π 2. We can cancel 5/(x – 2) from both the sides of
(1) to obtain x = 2
Thus, (1) has no solution.
and
Example 6: Suppose p ΠR. If the roots of 3x2 + 2x +
p(1 – p) = 0 are of opposite signs, then p must lie in the
interval
(a) (– •, 0)
(b) (0, 1)
(c) (1, •)
(d) (2, • )
Ans. (b)
Solution: (1 – p) + p(1 – p) + (1 – p) = 0
fi
(1 – p) [1 – p + p + 1] = 0 fi 2(1 – p) = 0
fi
p =1
Thus, equation becomes x2 + x = 0
fi
x = 0 or – 1
Example 7: If a, b are the roots of (x – a) (x – b) + c = 0,
c π 0, then roots of (ab – c) x2 + (a + b ) x + 1 = 0 are
(a) 1/a, 1/b
(b) – 1/a, – 1/b
(c) 1/a, –1/b
(d) –1/a, 1/b
Ans. (b)
Solution: a, b are roots of x2 – (a + b)x + ab + c = 0.
\
a + b = a + b, ab = ab + c
The equation
(ab – c) x2 + (a + b) x + 1 = 0
becomes
abx2 + (a + b)x + 1 = 0
fi
(ax + 1) (bx + 1) = 0
fi
x = – 1/a, – 1/b
Example 8: If a, b are roots x2 + px + q = 0, then value
of a3 + b3 is
(a) 3pq + p3
(b) 3pq – p3
(c) 3pq
(d) p3 – 3pq
Ans. (b)
Solution: a + b = – p, ab = q.
Now, a3 + b3 = (a + b)3 – 3ab(a + b )
= – p3 + 3pq = 3pq – p3
Example 9: If a ΠR and both the roots of
x2 – 6ax + 9a2 + 2a – 2 = 0 exceed 3, then a lies in the interval
(a) (1, •)
(b) (2, •)
(c) (11/9, •)
(d) f
Ans. (d)
Solution: Given equation can be written as (x – 3a)2 =
2(1 – a)
We must have 1 – a ≥ 0 or a £ 1.
Also,
x = 3a ± 2 1 - a.
Both the roots will exceed 3, if smaller of the two roots viz
3a - 2 1 - a exceeds 3, that is, 3a –
2 1- a > 3
(1)
fi
3(a – 1) > 2 1 - a.
For a = 1, this is not possible. For a < 1, we can write (1)
as
–3
which is no possible
(
2
1- a) > 2 1- a
Quadratic Equations 3.9
Thus, there is not real value of a.
7 - 4 3 , then x +
Example 10: If x =
(a) 2
(b) 3 7
(c) 4
Ans. (c)
(d) 4 7
1
is equal to:
x
2
Solution: 7 – 4 3 = 22 + ( 3 ) - 4 3
= (2 - 3 )
\
x=
and
2
Solution: For the equation to be defined, x ≥ 3, and
x2 + 7x + 10 ≥ 40 ⁄ x ≥ 3. Therefore, the given equation
has exactly one root, viz. 3.
Example 14: Suppose a ΠR. If 3x2 + 2(a2 + 1)x + (a2
– 3a + 2) = 0 possesses roots of opposite signs, then a lies
in the interval:
(a) (– •, –1)
(b) (–1, 1)
(c) (1, 2)
(d) (2, 3)
Ans. (c)
Solution: As the roots are of opposite signs, the product
of roots must be negative, that is,
7-4 3 =2- 3
1
1
=2+ 3
=
x
2- 3
1
=4
x
Example 11: Suppose a, b are roots of ax2 + bx + c = 0
then roots of a(x – 2)2 – b(x – 2) (x – 3) + c(x – 3)2 = 0 are
2+a 2+ b
2+a 3+a
,
,
(b)
(a)
1+ b 1+a
3+b 2+b
Thus, x +
2 + 3a 2 + 3b
,
(c)
2+a 2+ b
Ans. (d)
2 + 3a 2 + 3b
,
(d)
1+a 1+ b
fi
Example 15: Suppose a2 = 5a – 8 and b2 = 5b – 8, then
a
b
equation whose roots are and is
b
a
2
(a) 6x – 5x + 6 = 0
(b) 8x2 – 9x + 8 = 0
(c) 9x2 – 8x + 9 = 0
(d) 8x2 + 9x + 8 = 0
Ans. (b)
Solution: a, b are roots of x2 – 5x + 8 = 0.
\
a + b = 5, ab = 8
Now,
Solution: Write the equation as
2
Ê x - 2ˆ
Ê x - 2ˆ
+ bÁ a Á+c=0
˜
Ë x - 3 ˜¯
Ë x - 3¯
x-2
= a, b
x-3
2 + 3a 2 + 3b
,
fi
x=
1+a 1+ b
Example 12: If x, y, z ΠR then the least value of the
expression E = 3x2 + 5y2 + 4z2 – 6x + 20y – 8z – 3 is:
(a) – 15
(b) – 30
(c) – 45
(d) – 60
Ans. (b)
Now, –
Solution: We can write
E = 3(x2 – 2x + 1) + 5 (y2 + 4y + 4)
+ 4(z2 – 2z + 1) – 30
= 3(x – 1)2 + 5(y + 2)2 + 4(z – 1)2 – 30 ≥ –30
The least value is attained when
x = 1, y = – 2, z = 1.
Example 13: The number real of roots of
x - 3 (x2 + 7x + 10) = 0, where x ΠR is
(a) 0
(b) 1
(c) 2
(d) 3
Ans. (b)
a 2 - 3a + 2
< 0 fi (a – 1) (a – 2) < 0
3
a Π(1, 2)
a b
a 2 + b2 ( a + b )2 - 2 ab
+ =
=
b a
ab
ab
9
25
-2=
8
8
\ required equation is
x2 – (9/8)x + 1 = 0 or 8x2 – 9x + 8 = 0
=
Example 16: Sum of the roots of the equation
x2 + |2x – 3| – 4 = 0 is
(a) 2
(b) –2
(d) - 2
(c) 2
Ans. (c)
Solution: Case 1: When x < 3/2
In this case the equation becomes
x2 – (2x – 3) – 4 = 0
fi
x2 – 2x + 1 = 2
fi
fi
(x – 1)2 = 2 fi x – 1 = ±
x =1 ±
2
2
As x < 3/2, we take, x = 1 –
Case 2: When x ≥ 3/2
In this case the equation becomes
x2 + (2x – 3) – 4 = 0
fi
(x + 1)2 = 8 fi x + 1 ± 2 2
As x ≥ 3/2, x = –1 + 2 2
2
3.10
Complete Mathematics—JEE Main
Sum of the roots is
(1 –
2 ) + (– 1 + 2 2 ) =
Thus, f(x) = 0 has exactly one root in (–1, 1). Hence, ax2 +
bx + c = 0 has real and distinct roots.
2
Example 17: If the quadratic equation a(b – c) x2 +
b(c – a) x + c (a – b) = 0, where a, b, c are distinct real
numbers and abc π 0, has equal roots, then a, b, c are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.
Ans. (c)
Solution:
TIP
If sum of the coefficients of a polynomial equation is zero,
then 1 is a root of the equaiton.
As 1 is a root of the equation and equation has equal roots,
the other root must be 1. Thus we have
c (a - b)
a (b - c )
a(b – c) = c (a – b)
1 = product of roots =
fi
fi
b=
fi
Example 19: If both the roots of the quadratic equation
x2 + 2(a + 2) x + 9a – 1 = 0 are negative, then a lies in the set
(a) [1, •)
(b) (1/9, 1] » [4, •)
(c) (–2, 4) » (6, •)
(d) f
Ans. (b)
Solution: TIP: Both the roots are negative, if and only
if D ≥ 0, sum of roots < 0 and product of roots > 0
Thus, 4(a + 2)2 – 4(9a – 1) ≥ 0,
– 2 (a + 2) < 0, 9a – 1 > 0
fi
a2 – 5a + 4 ≥ 0, a > –2, a > 1/9
fi
(a – 1) (a – 4) ≥ 0, a > 1/9
fi
a £ 1 or a ≥ 4 and a > 1/9
fi
a Œ (1/9, 1] » [4, •)
Example 20: Suppose p ΠR and a, b are roots of
1
= 0, then the minimum value of a 4 + b 4 is
x2 – px +
2 p2
(a)
2ac
a+ c
a, b, c are in H.P.
(c)
Ans. (c)
Example 18: Suppose a, b, c Œ R, a π 0, and (a + c)2
< b . Then the quadratic equation ax2 + bx + c = 0 has
(a) real and distinct roots (b) real and equal roots
(c) purely imaginary roots (d) none of these
Ans. (a)
2
(b) 2 - 2
2 -2
(d) – 2
Solution: a + b = p, ab = 1/2p2
fi
a2 + b2 = (a + b)2 – 2ab = p2 – 1/p2
2
1 ˆ2
1
Ê
a4 + b 4 = (a2 + b2)2 – 2a2b2 = Á p2 - 2 ˜ - 4
Ë
2p
p ¯
\
= p4 +
Solution:
= p4 +
TIP
Suppose f(x) = ax2 + bx + c, a, b, c Œ R, a π 0. If f(a) f(b) <
0, then exactly one root of the equation f(x) = 0 lies between
a and b and the equation f(x) = 0 has real roots.
1
4
p
1
-2-
2 p4
1
2 p4
–2
1 ˆ
Ê
≥ 2 Á p4 ◊ 4 ˜
Ë
2p ¯
1/ 2
-2= 2 -2
[
Let f(x) = ax2 + bx + c.
Then f(1) = a + b + c, f(–1) = a – b + c.
We have f(1) f(–1) = (a + c)2 – b2 < 0
4
4
Thus, minimum possible value of a + b is
is attained when p = 1
AM ≥ GM]
2 - 2 , which
LEVEL 1
Straight Objective Type Questions
Example 21: The number of real solutions of
x+
is:
x +
(a) 0
(c) 2
Ans. (a)
x-2 =3
Solution: For
(1)
(b) 1
(d) 4
fi
x,
x - 2 to be defined x ≥ 0, x – 2 ≥ 0
x ≥ 2. But for x ≥ 2
x + x + x-2 ≥ 2 +
Thus, (1) has no real solution.
2 >3
Quadratic Equations 3.11
Example 22: In Fig. 3.16 graph of y = ax2 + bx + c is
given. Which one of the following is not true?
(a) a < 0
(b) c > 0
(c) b2 – 4ac > 0
(d) b < 0
Ans. (d)
Solution: We have
tan 25° + tan 20° = – 2p
and tan 25° tan 20° = q
Now,
1 = tan 45° = tan (25° + 20°)
fi
a
0
X
b
(a) tan a = c - a
b
Solution: The given parabola opens downwards, a < 0.
Also, y(0) = c > 0.
As roots of ax2 + bx + c = 0 are distinct, b2 – 4ac > 0.
However nothing can be said about b. It can be positive, zero
or negative. For instance, consider equations – x2 – 5x + 6 =
0, – x2 + 3x + 4 = 0 and – x2 + 9 = 0.
Example 23: Let a and b be the roots of px2 + qx + r = 0,
1 1
+
= 4, then value of
p π 0. If p, q, r are in A.P. and
a b
|a – b | is
1
2
(b)
(a)
61
17
9
9
1
34
9
(d)
2
13
9
Ans. (d)
Solution: 4 =
fi
Also,
Thus,
a +b
q
- q/ p
=
=ab
r/p
r
q + 4r = 0
2q = p + r
[ p, q, r are in A.P.]
p = – 9r, q = – 4r
|a – b|2 = (a + b)2 – 4ab
- q 2 4r
= ÊÁ ˆ˜ Ë p¯
p
=
fi
|a – b| =
1 – q = – 2p fi 2p – q = – 1
Example 25: Suppose 0 < a < b, and 2a + b = p/2. If tan a,
tan b are roots of ax2 + bx + c = 0, then
Fig. 3.16
(c)
tan(25∞) + tan(20∞)
-2 p
=
1 - tan(25∞) tan(20∞) 1 - q
=
Y
(c) tan a =
(b) tan b = c + a
b
b
c-a
(d) tan b = –
b
c+a
Ans. (a)
Solution: tan a + tan b = – b/a and tan a tan b = c/a.
tan a + tan b
Now,
tan (a + b ) =
1 - tan a tan b
fi
tan (p/2 – a) =
fi
cot a =
- b/a
1- c/a
-b
a-c
c-a
.
b
Example 26: The value of a for which one root of the
equation x2 – (a + 1) x + a2 + a – 8 = 0 exceeds 2 and other
is less than 2, are given by
(a) 3 < a < 10
(b) a ≥ 10
(c) – 2 < a < 3
(d) a £ – 2
Ans. (c)
Thus,
tan a =
Solution: As the roots are real and distinct
(a + 1)2 – 4(a2 + a – 8) > 0
3a2 + 2a – 33 < 0
11
< a < 3 (1)
fi
(3a + 11) (a – 3) < 0 fi 3
Y
16
Ê 1 ˆ 52
- 4Ë - ¯ =
81
9
81
2
13
9
Example 24: If tan 25° and tan 20° are roots of the
quadratic equation x2 + 2px + q = 0, then 2p – q is equal to
(a) – 2
(b) – 1
(c) 0
(d) 1
Ans. (b)
0
2
Also, for
22 – 2(a + 1) + a2 + a – 8 < 0
fi
a2 – a – 6 < 0 fi (a – 3) (a + 2) < 0
fi
–2<a<3
From (1) and (2), we get – 2 < a < 3.
(2)
3.12
Complete Mathematics—JEE Main
Example 27: The least integral value a of x such that
x-5
> 0 , satisfies the relation:
2
x + 5 x - 14
(a) a2 + 3a – 4 = 0
(b) a2 – 5a + 4 = 0
2
(c) a – 7a + 6 = 0
(d) a2 + 5a – 6 = 0
Ans. (d)
x-5
x-5
= E(say)
2
(
x
7) ( x - 2 )
+
x + 5 x - 14
Sign of E in different intervals are shown below
Solution: 0 <
Sign of
E
-
=
+
-7
+
2
5
x-5
>0
( x + 7) ( x - 2 )
fi
– 7 < x < 2 or x > 5.
Therefore the least integral value a of x is – 6. This value
of a satisfies the relation a2 + 5a – 6 = 0.
Thus,
Example 28: Let p and q be two non-zero real numbers
and a, b are two numbers such that a3 + b3 = – p, ab = q,
then the quadratic equation whose roots are a 2/b and b 2/a is
(b) qx2 + px + q2 = 0
(a) px2 – qx + p2 = 0
(c) px2 + qx + p2 = 0
(d) qx2 – px + q2 = 0
Ans. (b)
Solution:
and
a2 b2 a3 + b3 -p
+
=
=
q
b
a
ab
Ê a2 ˆ Ê b2 ˆ
ÁË b ˜¯ ÁË a ˜¯ = ab = q
Thus, required equation is
p
x2 – ÊÁ - ˆ˜ x + q = 0
Ë q¯
qx2 + px + q2 = 0.
or
Example 29: The set of values of a for which the
quadratic equation
(a + 2) x2 – 2ax – a = 0
has two roots on the number line symmetrically placed
about the point 1 is
(a) {– 1, 0}
(b) {0, 2}
(c) f
(d) {0, 1}
Ans. (c)
Solution: Let roots of the equation be
1 – k and 1 + k, where k > 0
Then
2 = (1 – k) + (1 + k) =
2a
a
fi1=
a +2
a +2
Example 30: Suppose k ΠR and the quadratic equation
x2 – (k – 3)x + k = 0 has at least one positive roots, then k lies
in the set:
(a) (– •, 0) » [9, 16] (b) (– •, 0) » [16, 8)
(c) (– •, 0) » [9, •)
(d) (– •, 0) » (1, •)
Ans. (c)
Solution: Both the roots a, b will be non-positive if
D ≥ 0, a + b £ 0, ab ≥ 0
fi
(k – 3)2 – 4k ≥ 0, (k – 3) £ 0, k ≥ 0
fi
(k – 1) (k – 9) ≥ 0, k £ 3, k ≥ 0
fi
0 £ k £ 1.
Thus, quadratic equation will have at least one positive
root if k < 0 or k > 1 and (k £ 1 or k ≥ 9)
fi
k Œ (– •, 0) » [9, •)
Example 31: Two non-integer roots of the equation
(1)
(x2 + 3x)2 – (x2 + 3x) – 6 = 0
are
(a)
1
1
(- 3 + 11), (- 3 - 11)
2
2
(b)
1
1
(- 3 + 7 ), (- 3 - 7 )
2
2
1
1
(- 3 + 21), (- 3 - 21)
2
2
(d) none of these
Ans. (c)
(c)
Solution: TIP: It is an equation which is reducible to
quadratic. Put x2 + 3x = y.
The equation (1) becomes
y2 – y – 6 = 0 fi (y + 2) (y – 3) = 0
fi
y = – 2, y = 3
When
y = 2,
x2 + 3x = – 2 fi x2 + 3x + 2 = 0
fi (x + 1) (x + 2) = 0 fi x = – 1, – 2
When
y = 3, x2 + 3x = 3 fi x2 + 3x – 3 = 0
1
1
fi
x=
(- 3 ± 9 + 12 ) = (- 3 ± 21)
2
2
Example 32: Two non-integer roots of
4
are
2
Ê 3x - 1 ˆ
Ê 3x - 1 ˆ
ÁË 2 x + 3 ˜¯ – 5 ÁË 2 x + 3 ˜¯ + 4 = 0
(a) – 5/7, – 2/5
(c) 5/7, 7/5
Ans. (a)
(b) – 2/5, 7/5
(d) – 2/5, 3/5
Solution: TIP: It is an equation which is reducible to
2
This is not possible for any value of a.
(1)
Ê 3x - 1 ˆ
quadratic. Put Á
= y. The equation (1) becomes
Ë 2 x + 3 ˜¯
Quadratic Equations 3.13
y2 – 5y + 4 = 0 fi (y – 1) (y – 4) = 0 fi y = 1, 4
Solution: Put x +
2
When
3x - 1
Ê 3x - 1 ˆ
=1fi
=±1
y = 1, Á
Ë 2 x + 3 ˜¯
2x + 3
fi
x = 4, – 2/5
When
3x - 1
Ê 3x - 1ˆ
y = 4, Á
=4fi
=±2
˜
Ë 2 x + 3¯
2x + 3
1
1
= y, so that x2 +
= y2 – 2. Then
x
x
equation (1) becomes
2(y2 – 2) – 9y + 14 = 0
fi
2y2 – 9y + 10 = 0 fi
fi
y = 5/2, 2
2
(2y – 5) (y – 2) = 0
fi
x = – 7, – 5/7
Thus, required roots are – 2/5, – 5/7.
When x +
1
5
1
= , we get x = 2,
x
2
2
Example 33: Sum of the roots of the equation
is
4x – 3(2x + 3) + 128 = 0
(a) 5
(b) 6
(c) 7
(d) 8
Ans. (c)
When x +
1
= 2, we get x = 1.
x
(1)
Example 36: The non-integral roots of
x4 – 3x3 – 2x2 + 3x + 1 = 0
are
Example 34: The only real value of x satisfying the
equation is
where x ΠR
(a) 4/35
(c) 16/3
Ans. (d)
Solution: For (1) to be defined either x < – 4 or x > 0.
x
= y. Equation (1) becomes
Put
x+4
fi
As
Thus,
fi
(6y + 1) (y – 2) = 0 fi
y=
6y2 – 11y – 2 = 0
y = – 1/6, y = 2.
x
1
≥ 0, we reject y = – .
6
x+4
x
= y2 = 4
x+4
x = 4x + 16 fi x = – 16/3 < – 4
y =2
Ans. (a)
Solution: Dividing (1) by x2, we get
1
3
+ 2 =0
x2 – 3x – 2 +
x
x
(1)
(b) – 4/35
(d) none of these.
6y – 2/y = 11 fi
1
1
(3 + 13 ), (3 - 13 )
2
2
1
1
(3 - 13 ), (- 3 - 13 )
(b)
2
2
1
1
(3 + 17 ), ( 3 - 17 )
(c)
2
2
(d) none of these.
(a)
Solution: Put 2x = y. Equation (1) becomes
y2 – 3(8y) + 128 = 0 fi y2 – 24y + 128 = 0
fi
(y – 8) (y – 16) = 0 fi y = 16, 8
fi
2x = 16, 8 fi x = 4, 3
\ Sum of the roots is 7.
x
x+4
6
-2
= 11
x+4
x
(1)
fi
Example 35: The number of real values of x satisfying
the equation
1ˆ
1ˆ
Ê
Ê
2 Á x 2 + 2 ˜ – 9 Á x + ˜ + 14 = 0
(1)
Ë
Ë
x¯
x ¯
is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (c)
1ˆ
Ê
Ê 2 1ˆ
ÁË x + 2 ˜¯ – 3 ÁË x - ˜¯ – 2 = 0
x
x
Put x – 1/x = y, so that
1
1
x2 + 2 – 2 = y2 or x2 + 2 = y2 + 2
x
x
Equation (2) now becomes
y2 + 2 – 3y – 2 = 0 fi y2 – 3y = 0
fi
y(y – 3) = 0 fi y = 0 or y = 3
or
When y = 0, we get x –
1
=0
x
When y = 3, we get x –
1
=3
x
fi
x2 – 3x – 1 = 0
fi
fi
(2)
x=±1
x=
1
(3 ± 13 )
2
Example 37: Suppose a, b, c Œ R and b π c. If a, b are
roots of x2 + ax + b = 0 and g, d are roots of x2 + ax + c = 0,
then value of
(a - g ) (a - d )
is independent of:
(b - g ) (b - d )
(a) a, b
(b) b, c
(c) c, a
(d) a, b, c
3.14
Complete Mathematics—JEE Main
Solution: Note that we must have 3x2 + x + 5 ≥ 0 and
x – 3 ≥ 0 or x ≥ 3. Squaring both sides of (1), we get
3x2 + x + 5 = x2 – 6x + 9
Ans. (d)
Solution: x2 + ax + c = (x – g ) (x – d )
Thus,
=
-b+c
=1
-b+c
Example 38: Number of real solutions of
(x – 1) (x + 1) (2x + 1) (2x – 3) = 15
(1)
is
(a) 0
(c) 3
Ans. (b)
(b) 2
(d) 4
fi
When y = 6, 2x – x = 6
fi (y – 6) (y + 2) = 0
fi
2
fi
2x – 4x + 3x – 6 = 0
fi
2x – x – 6 = 0
(2x + 3) (x – 2) = 0
x = – 3/2, 2
When y = – 2, 2x2 – x = – 2
or
2x2 – x + 2 = 0
As D = 1 – 4 (2) (2) = – 15 < 0, we get 2x2 – x + 2 = 0 does
not have real roots.
Example 39: If p, q ŒR and 2 + 3i a root of x2 + px +
q = 0, then
(b) p = – 4, q = 7
(a) p = – 2, q = 3
(c) p = 3, q = 2
(d) p = – 4, q = 2
Ans. (b)
Solution: Other root of the equation is 2 –
– p = (2 + 3i ) + (2 – 3i )
and
fi
3i . Thus
(2)
q = (2 + 3i )(2 – 3i )
p = – 4, q = 7
Example 40: The number of solutions of
3 x 2 + x + 5 = x Р3, where x ΠR
is:
(a) 0
(c) 2
Ans. (a)
(b) 1
(d) 4
is:
(a) 0
(c) 2
Ans. (c)
(b) 1
(d) 3
x+9 =5–
4-x
x + 9 = 25 – 10 4 - x + 4 – x
fi
10 4 - x = 20 – 2x fi 5 4 - x = 10 – x
Squaring both the sides we get
25 (4 – x) = 100 – 20x + x2 fi x2 + 5x = 0
fi
x(x + 5) = 0 fi x = 0 or x = – 5
Both x = 0 and x = – 5 satisfy (1).
2
fi
(1)
Squaring both the sides we get
y = 6, – 2.
2
Example 41: The number of solutions of
4-x + x+9 =5
We can write (1) as
fi
(2x2 – x – 1) (2x2 – x – 3) = 15
(2)
2
Put 2x – x = y, so that (2) becomes
(y – 1) (y – 3) = 15 fi
y2 – 4y + 3 = 15
y2 – 4y – 12 = 0
(2x – 1) (x + 4) = 0
Solution: Note that 4 – x ≥ 0 and x + 9 ≥ 0 fi – 9 £ x £ 4.
Solution: We write (1) as
(x – 1) (2x + 1) (x + 1) (2x – 3) = 15
fi
fi
fi
x = 1/2, – 4
None of these satisfy the inequality x ≥ 3. Thus, (1) has no
solution.
a, b are roots of x2 + ax + b = 0]
[
2x2 + 7x – 4 = 0
fi
a 2 + aa + c
(a - g ) (a - d )
= 2
b + ab + c
(b - g ) (b - d )
(1)
Example 42: Suppose a Œ R and a, b are roots of x2 –
4ax + 5a2 – 6a = 0. The maximum possible distance between
a and b is
(a) 6
(b) 5
(c) 3
(d) 1
Ans. (a)
Solution: (a – b)2 = (a + b)2 – 4 ab
= 16a2 – 4(5a2 – 6a)
= 4(6a – a2)
= 4[9 – (3 – a)2]
As a ΠR, (a Рb)2 is maximum when a = 3. Thus, maximum
possible distance between a and b is 6.
Example 43: The value of a for which one root of the
quadratic equation.
(1)
(a2 – 5a + 3) x2 + (3a – 1)x + 2 = 0
is twice the other, is
(a) – 2/3
(b) 1/3
(c) – 1/3
(d) 2/3
Ans. (d)
Solution: Let a and 2a be the roots of (1), then
(a2 – 5a + 3) a2 + (3a – 1)a + 2 = 0
2
2
(2)
and (a – 5a + 3) (4a ) + (3a – 1) (2a) + 2 = 0
(3)
Multiplying (2) by 4 and subtracting it form (3) we get
Quadratic Equations 3.15
(3a – 1) (2a) + 6 = 0
E = (c – a) (c – b) (– d – a) (– d – b)
= (c2 + pc + 1) [(– d)2 – pd + 1]
[∵ a + b = – p]
2
2
= (c + pc + 1) (d – pd + 1)
Clearly a π 1/3. Therefore,
a = – 3/(3a – 1)
Putting this value in (2) we get
(a2 – 5a + 3) (9) – (3a – 1)2 (3) + 2(3a – 1)2 = 0
fi
9a2 – 45a + 27 – (9a2 – 6a + 1) = 0
fi
– 39a + 26 = 0
But c2 + qc + 1 = 0 and d 2 + qd + 1 = 0
\
fi
a = 2/3.
2
For a = 2/3, the equation becomes x + 9x + 18 = 0, whose
roots are – 3, – 6.
x2 + x + 2
Example 44: Range of function f(x) = 2
,
x + x +1
x ΠR is
(a) (1, •)
(b) (1, 3/2)
(c) (1, 7/3]
(d) (1, 7/5]
Ans. (c)
Solution: Let y =
x2 + x + 2
x2 + x + 1
=1+
=1+
1
= x2 + x + 1 =
y -1
fi
1
x2 + x + 1
1
( x + 1/ 2 ) 2 + 3 / 4
Ê
ÁË x +
g(x) = – x2 – 2cx + b2 = b2 + c2 – (x + c)2
fi
2
c > 2b fi |c| >
3
x
3/2
2
4
8
Ê 2 ˆ
Ê 4ˆ
Ê 4ˆ
=
= Á 1/ 2 ˜ = Á ˜
ÁË ˜¯ = ¥
Ë 3¯
3
3
3 3 3 Ë3 ¯
x = 3/2
Example 48: If x2 + 2ax + 10 Р3a > 0 for each x ΠR,
then
(a) a < –5
(b) – 5 < a < 2
(c) a > 5
(d) 2 < a < 5
Ans. (b)
fi
fi
fi
fi
Solution: x2 + 2ax + 10 – 3a > 0 ⁄ x Œ R
(x + a)2 – (a2 + 10 – 3a) > 0 ⁄ x Œ R
a2 + 3a – 10 < 0
(a + 5) (a – 2) < 0
–5<a<2
Example 49: The number of solutions of
2
2
As min f(x) > max g(x), we get 2c – b > b + c
2
3x
4x
= 3x 3 +
2
3
Ê 3 ˆ 4x = Ê 4 ˆ 3x
ÁË ˜¯
ÁË ˜¯
2
3
4x +
fi
max g(x) = b2 + c2
2
fi
Solution: 4x – 3x – 1/2 = 3x + 1/2 – 22x – 1
fi
4x + 22x – 1 = 3x + 1/2 + 3x – 1/2
fi
Solution: f(x) = (x + b)2 + 2c2 – b2
fi
min f(x) = 2c2 – b2
[∵ cd = 1]
Example 47: If 4x – 3x – 1/2 = 3x + 1/2 – 22x – 1, then value
of x is equal to
(a) 5/2
(b) 2
(c) 3/2
(d) 1
Ans. (c)
2
Example 45: If f(x) = x2 + 2bx + 2c2 and
g(x) = – x2 – 2cx + b2 are such that min f(x) > max g(x), then
relation between b and c, is:
(a) no relation
(b) 0 < c < b/2
|b|
(d) |c| > 2 |b|.
(c) |c| <
2
Ans. (d)
fi
= cd (q2 – p2) = q2 – p2
>1
Thus, y > 1 and y – 1 £ 4/3 fi y £ 7/3
\
1 < y £ 7/3
Also
= cd(q – p) (q + p)
fi
1ˆ
3 3
˜¯ + ≥
2
4 4
E = (– qc + pc) (– qd – pd)
2
2 |b|
Example 46: If a, b are the roots of x2 + px + 1 = 0, and
c, d are the roots of x2 + qx + 1 = 0, then the value of
E = (a – c) (b – c) (a + d) (b + d)
is:
(a) p2 – q2
(b) q2 – p2
2
2
(c) q + p
(d) none of these
Ans. (b)
Solution: We have
x2 + px + 1 = (x – a) (x – b)
Thus,
x +1 - x -1 = 1
(x ΠR)
is
(a) 1
(c) 4
Ans. (a)
(b) 2
(d) infinite
Solution: For the equation to make sense we must have
x + 1 ≥ 0 and x – 1 ≥ 0 fi x ≥ – 1, x ≥ 1 i.e. x ≥ 1.
We rewrite equation as
x +1 =1+ x -1
and square both the sides to obtain
x + 1 = 1 + x – 1 + 2 x -1
3.16
fi
Complete Mathematics—JEE Main
1
1
5
= x -1 fi
= x – 1 or x =
2
4
4
Also, x = 5/4 satisfies the given equation.
Example 50: If a, b, c are real and a π b, then the roots
of the equation
(1)
2(a – b)x2 – 11(a + b + c)x – 3(a – b) = 0
are
(a) real and equal
(b) real and unequal
(c) purely imaginary (d) none of these
Ans. (b)
Solution: The discriminant D of the quadratic equation
(1) is given by
D = 121(a + b + c)2 + 24(a – b)2
As a, b, c are real, 121(a + b + c)2 ≥ 0
Also, as a π b, (a – b)2 > 0
Thus, D > 0.
Therefore, equation (1) has real and unequal roots.
Example 51: Let a > 0, b > 0 and c > 0. Then both the
roots of the equation
2ax2 + 3bx + 5c = 0
(1)
(a) are real and negative
(b) have negative real parts
(c) have positive real parts
(d) none of these
Ans. (b)
Solution: The discriminant D of the equation (1) is
given by
D = (3b)2 – 4(2a) (5c) = 9b2 – 40ac
If D ≥ 0, then both the roots of (1) are real.
Also, since ac > 0, D < 9b2.
D < 3b [∵ b > 0]
fi
Therefore, in this case both the roots
- 3b - D
and
4a
- 3b + D
are negative.
4a
If D < 0, both the roots of (1) are imaginary and are given by
- 3b ± i 40ac - 9b2
2
Both these roots have negative real parts.
x=
Example 52: If a, b, c are real, then both the roots of
the equation
(x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0
(1)
are always
(a) positive
(b) negative
(c) real
(d) none of these
Ans. (c)
Solution: We can write (1) as
3x2 – 2(a + b + c)x + bc + ca + ab = 0
(2)
The discriminant D of (2) is given by
D = 4(a + b + c)2 – 4 ¥ 3(bc + ca + ab)
= 4[a2 + b2 + c2 + 2bc + 2ca + 2ab – 3bc – 3ca – 3ab]
= 4[a2 + b2 + c2 – bc – ca – ab]
= 2[(b2 + c2 – 2bc) + (c2 + a2 – 2ca) + (a2 + b2 – 2ab)]
= 2[(b – c)2 + (c – a)2 + (a – b)2]
As a, b, c are real, we get D ≥ 0. Thus, roots of (1) are real.
Example 53: If the roots of the equation (4a – a2 – 5) x2 –
(2a Р1)x + 3a = 0, a ΠR, are real and lie on opposite sides
of unity, then the a lies in the interval:
(a) (1, 5)
(b) (1, 4)
(c) (3, •)
(d) (– •, 4)
Ans. (b)
Solution: TIP: If roots of a quadratic equation f(x) = 0
are on the opposite sides of a real number p, then f(x + p) =
0 has roots of opposite sings.
Put
y = x – 1, so that x = y + 1
The equation now becomes:
(4a – a2 – 5) (y + 1)2 – (2a – 1) (y + 1) + 3a = 0
fi (4a – a2 – 5)y2 + (8a – 2a2 – 10 – 2a + 1)y
+ (4a – a2 – 5 – 2a + 1 + 3a) = 0
fi (4a – a2 – 5)y2 + (6a – 2a2 – 9)y
+ (5a – a2 – 4) = 0
This equation must have roots of opposite signs. This is
possible if and only if
5a - a 2 - 4
4a - a2 - 5
< 0 or
a 2 - 5a + 4
a 2 - 4a + 5
<0
As a2 – 4a + 5 = (a – 2)2 + 1 > 0 ⁄ a,
we get (a – 1) (a – 4) < 0 fi 1 < a < 4.
Example 54: If a, b are roots of ax2 + bx + c = 0, then
roots of
a3x2 + abcx + c3 = 0 are
(a) ab, a + b
(c) ab, a2b2
Ans. (b)
(b) a2b, ab2
(d) a3, b3
–b
c
, ab = .
a
a
3 2
3
as
a x + abcx + c = 0
Solution: a + b =
Write
fi
c 3
x +Ê ˆ = 0
Ë a¯
a
2
x – (a + b) abx + a3b3 = 0
fi
fi
fi
x2 – (a2b + ab2)x + (a2b) (ab2) = 0
(x – a2b) (x – ab2) = 0
x = a2b, ab2
x2 +
bc
2
Quadratic Equations 3.17
Example 55: If P(x) = ax2 + bx + c and Q(x) = – ax2 + dx
+ c, where a, b, c ΠR, then P(x) Q(x) = 0 has
(a) no real root
(b) exactly two real roots
(c) at least two real roots
(d) none of these
Ans. (c)
Solution: Let D1 = b2 – 4ac and D2 = d2 + 4ac
We have D1 + D2 = b2 + d2 ≥ 0
fi at least one of D1, D2 ≥ 0
fi one of P(x) = 0 or Q(x) = 0
has real roots. Thus, P(x) Q(x) = 0
has at least two real roots.
Example 56: If the product of the roots of the equation
x2 – 5kx + 2e4lnk – 1 = 0 is 31, then sum of the root is
(a) – 10
(b) 5
(c) – 8
(d) 10
Ans. (d)
Solution: We have product of the roots
= 2e4lnk – 1 = 31 (given)
fi
e4lnk = 16
fi
k4 = 16 fi k4 – 16 = 0
fi
(k – 2) (k + 2) (k2 + 4) = 0
fi
k = 2, – 2
As k > 0, we get k = 2.
Sum of the roots = 5k = 10
Example 57: Let a, b be the roots of the equation x2 –
a
px + r = 0 and , 2b be the roots of the equation x2 – qx + r
2
= 0. Then the value of r is :
2
2
(a)
(p – q) (2q – p) (b)
(q – p) (2p – q)
9
9
2
2
(q – 2p) (2q – p) (d)
(2p – q) (2q – p)
(c)
9
9
Ans. (d)
Solution: a + b = p, ab = r
a
+ 2b = q.
and
2
But
a + b = p and a + 4b = 2q
1
2
fi
b = (2q – p) and a = (2p – q)
3
3
Thus,
ab = r
2
(2p – q) (2q – p) = r
fi
9
Example 58: Sum of all the values of x satisfying the
equation
log17 log11 x + 11 + x = 0
(1)
is:
(a) 25
(b) 36
(c) 171
(d) 0
(
)
Ans. (a)
Solution: Equation (1) is defined if x ≥ 0.
We can rewrite (1) as
log11 x + 11 + x = 170 = 1
1
fi
x + 11 + x = 11 = 11
fi
x + 11 = 11 – x
Squaring both the sides we get
x + 11 = 121 – 22 x + x
(
)
fi
22 x = 110 fi x = 5 or x = 25
This clearly satisfies (1).
Thus, sum of all the values satisfying (1) is 25.
1
1
and
are the roots of the
a
b
equation ax2 + bx + 1 = 0 (a π 0, a, b, Œ R), then the equation,
x(x + b3) + (a3 – 3abx) = 0 has roots:
(a) a3/2 and b3/2
(b) ab–3/2 and a1/2 b
Example 59: If
(c)
Ans. (a)
(d) a–3/2 and b –3/2
ab and a b
Solution: The equation x(x + b3) + (a3 – 3abx) = 0 can
be written as
x2 + (b3 – 3ab)x + a3 = 0
We have
1
1
b
1
1
+
=
=– ,
a a b a
a
b
a + b = - b, a b = a
Now
– (b3 – 3ab) = (–b)3 – 3a(–b)
=
(
3
a + b) -3( a + b) a b
= ( a )3 + ( b )3 = a 3 2 + b 3 2
and a3 = a3/2 b3/2
\ The roots of the equation
x2 + (b2 – 3ab)x + a3 = 0 are a3/2, b3/2
Example 60: If p, q are roots of x2 + px + q = 0, then
(a) p = 1
(b) p = 1 or 0
(c) p = – 2
(d) p = – 2 or 0
Ans. (b)
Solution: We have p + q = – p, pq = q
As
pq = q,
we get
q(p – 1) = 0
fi
q = 0 or p = 1
If q = 0, we get p = 0
If p = 1, we get p + q = – p fi q = – 2.
Thus, p = 1 or p = 0.
Example 61: The equation
x + 1 - x - 1 = 4 x - 1 , (x ΠR)
3.18
Complete Mathematics—JEE Main
(a)
(b)
(c)
(d)
Ans. (a)
no solution
one solution
two solutions
more than two solutions
Solution: The given equation is valid if x + 1 ≥ 0, x – 1
≥ 0 and 4x – 1 ≥ 0 i.e. if x ≥ 1.
Squaring both the sides we get
x + 1 + x – 1 – 2 ( x + 1) ( x - 1) = 4x – 1
fi
1 – 2x = 2 ( x + 1) ( x - 1)
Squaring again, we get
1 – 4x + 4x2 = 4(x2 – 1)
fi
4x = 5 or x = 5/4.
Putting this value of x in the given equation, we get
5
5
Ê 5ˆ
+1 - 1 = 4Á ˜ - 1
Ë 4¯
4
4
3 1
- = 2 or 1 = 2
2 2
fi
which is not true.
Thus, the given equation has no solution.
Example 62: The sum of all the real roots of the equation
(1)
|x – 2|2 + |x – 2| – 2 = 0
is
(a) 7
(c) 1
Ans. (b)
fi
As
\
fi
fi
\
(b) 4
(d) none of these
Solution: Putting y = |x – 2|, we can rewrite (1) as
y2 + y – 2 = 0 or y2 + 2y – y – 2 = 0
y(y + 2) – (y + 2) = 0 fi (y – 1) (y + 2) = 0
y = |x – 2| ≥ 0, y + 2 ≥ 2
y–1=0fiy=1
|x – 2| = 1 fi x – 2 = ± 1
x = 2 ± 1 fi x = 3 or 1
Sum of the roots = 4
Example 63: Let p and q be the roots of x2 – 2x + A = 0
and let r and s be the roots of x2 – 18x + B = 0. If p < q < r <
s are in A.P. then ordered pair (A, B) is equal to
(a) (– 3, 77)
(b) (77, – 3)
(c) (– 3, – 77)
(d) none of these
Ans. (a)
Solution: We have p + q = 2, pq = A
(1)
and
r + s = 18, rs = B
(2)
As p, q, r, s are in AP, we take
p = a – 3d, q = a – d, r = a + d, s = a + 3d.
As p < q < r < s, we have d > 0
Now,
2 = p + q = 2a – 4d
and
18 = r + s = 2a + 4d
Solving above equations, we get a = 5 and d = 2
\
p = – 1, q = 3, r = 7 and s = 11
Thus,
A = pq = – 3 and B = rs = 77.
Example 64: In a triangle PQR, –R = p/2. If tan (P/2)
and tan (Q/2) are the roots of the equation ax2 + bx + c = 0
where a π 0, then
(a) a + b = c
(b) b + c = a
(c) a + c = b
(d) b = c
Ans. (a)
Solution: We have
-b
Ê Pˆ
Ê Qˆ
tan Á ˜ + tan Á ˜ =
Ë 2¯
Ë 2¯
a
P
Q
c
Ê ˆ
Ê ˆ
and
tan Á ˜ tan Á ˜ =
Ë 2¯
Ë 2¯
a
Now,
P+Q=
=
fi
fi
p
Êpˆ
Ê P Qˆ
fi 1 = tan Á ˜ = tan Á + ˜
Ë
¯
Ë 2 2¯
2
4
tan ( P / 2 ) + tan (Q / 2 )
1 - tan ( P / 2 ) tan (Q / 2 )
- b/a
-b
=
1 - c/a a - c
a – c =– b fi c = a + b
1=
Example 65: If a and b (a < b), are the roots of the
equation x2 + bx + c = 0 where c < 0 < b, then
(a) 0 < a < b
(b) a < 0 < b < |a |
(c) a < b < 0
(d) a < 0 < |a | < b
Ans. (b)
Solution: We have a + b = – b, ab = c
As c < 0, b > 0, we get a < 0 < b
Also,
b = – b – a < – a = |a |
Thus,
a < 0 < b < |a |
Example 66: For the equation 3x2 + px + 3 = 0, p > 0, if
one of the roots is square of the other, than p is equal to
(a) 1/3
(b) 1
(c) 3
(d) 2/3
Ans. (c)
Solution: Let a, a2 be the roots of 3x2 + px + 3 = 0.
Then a + a2 = – p/3 and a ◊ a2 = 1
As
a3 = 1, we get a = 1, w or w2.
If
a = 1, then p = – 3(a + a2) = – 6 < 0.
Not possible as p > 0.
Thus, a π 1.
-p
We take a = w and a2 = w2, so that
= w + w2 = – 1
3
fi
p =3
Example 67: Let a, b be roots of ax2 + bx + c = 0, where
ac π 0. Roots of cx2 – bx + a = 0 are
Quadratic Equations 3.19
(a) 1/a, 1/b
(c) 1 / a , 1 / b
Ans. (b)
(b) –1/a, – 1/b
(d) -1 / a , - 1 / b
When
b
ba
a+b
=–
=c
ab
ca
a(1 + r) = 1 fi a = 1/3
In this case
a
1 Ê 1ˆ Ê 1ˆ
=
= c
ab Ë a ¯ ÁË b ˜¯
Thus, roots of cx2 – bx + a = 0 are - 1 a , - 1 b .
Example 68: If b > a, then the equation (x – a) (x – b)
– 1 = 0 has
(a) both roots in [a, b]
(b) both roots in (– •, a)
(c) both roots in (b, •)
(d) one root in (– •, a) and other in (b, •).
Ans. (d)
Solution: Graph of y = (x – a) (x – b) – 1 is given in
Fig. 3.17.
It is a parabola which open upwards. Also, y < 0 for x = a
and x = b.
\ y = (x – a) (x – b) – 1 meets the x-axis at two points
once in (– •, a) and once in (b, •). Thus, one root lies in
(– •, a) and one in (b, •).
and
q = gd = (ar2) (ar3) = a 2 r 5 = – 32
Hence,
p = – 2, q = – 32.
Example 70: If a, b, and c are not all equal and a and b
be the roots of the equation ax2 + bx + c = 0, then value of
(1 + a + a2) (1 + b + b2) is
(a) 0
(b) positive
(c) negative
(d) non-negative
Ans. (d)
Solution: We have a + b = – b/a, ab = c/a
Now, (1 + a + a2) (1 + b + b2)
= 1 + a + b + a 2 + b 2 + ab + a 2b + ab 2 + a 2b 2
= 1 + (a + b) + (a + b)2 – 2ab + ab [1 + a + b + ab ]
= 1=
b
x axis
p = ab = a(ar) = a2r
1
2
= (2 ) =
9
9
which is not an integer.
Thus, r = – 2. In this case, a (1 + r) = 1 fi a = – 1.
\
p = a 2r = (– 1)2 (– 2) = – 2
y
a
a r 2 (1 + r )
= 4 fi r2 = 4 fi r = ± 2
a (1 + r )
r = 2,
we get
Ê 1ˆ Ê 1ˆ
= Ë- ¯ + Á- ˜
Ë b¯
a
And
ar2 (1 + r) = 4
fi
Thus,
Solution: a + b = – b/a, ab = c/a.
Now,
g + d = 4 fi ar2 + ar3 = 4
=
b b 2 2c c c
+
+ +
a a2
a a a
1
2
a
1
2a2
2
Ê bˆ c
+
ÁË
˜
a ¯ a2
{a2 + b2 + c2 – bc – ca – ab}
[(b – c)2 + (c – a)2 + (a – b)2] ≥ 0
Example 71: If a, b, c are in A.P. and if the equations
(1)
(b – c)x2 + (c – a)x + (a – b) = 0
-1
Fig. 3.17
2
Example 69: Let a, b be the roots of x – x + p = 0 and g,
d be the roots of x2 – 4x + q = 0. If a, b, g, d are in G.P. then
the integral values of p and q respectively, are
(a) – 2, – 32
(b) – 2, 3
(c) – 6, 3
(d) – 6, – 32
Ans. (a)
Solution: We have
a + b = 1, ab = p,
g + d = 4, gd = q
Let r be the common ratio of the GP a, b, g, d. Then
a + b = 1 fi a + ar = 1 fi a(1 + r) = 1
and
2(c + a)x2 + (b + c)x = 0
have a common root, then
(a) a2, b2, c2 are in A.P.
(b) a2, c2, b2 are in A.P.
(c) c2, a2, b2 are in A.P.
(d) none of these
Ans. (b)
(2)
Solution: Clearly x = 1 is a root of (1). If a is the other
root of (1), then
a = 1◊a =
=1 [
a-b
[product of roots]
b-c
a, b, c are in A. P.]
3.20
Complete Mathematics—JEE Main
Thus, the roots of (1) are 1, 1. Now, (1) and (2) will have
a common root if 1 is also a root of (2).
fi 2(c + a) + b + c = 0
fi 2(2b) + b + c = 0 fi c = – 5b
Also,
a + c = 2b
fi
[∵ a, b, c are in AP]
a = 2b – c = 2b + 5b = 7b
\
a2 = 49b2, c2 = 25b2
This, show that a2, c2, b2 are in A.P.
a b
a 2 + b 2 (a + b )2
+
=
=
–2
b a
ab
ab
25
19
=
–2=
3
3
Êaˆ Ê bˆ
and
ÁË b ˜¯ Ë ¯ = 1.
a
a b
Thus, the quadratic equation whose roots are ,
is
b a
Next,
x2 –
Example 72: Value of
6 + 6 + 6 + up to •
x=
(1)
is
(a) 3
(c) 1
Ans. (a)
(b) 2
(d) none of these
Solution: We can write (1) as
x = 6 + x fi x2 = 6 + x
fi
x2 – x – 6 = 0 fi (x – 3) (x + 2) = 0
fi
x = 3, – 2
As x > 0, we take x = 3
Example 73: Two complex numbers a and b are such
that a + b = 2 and a4 + b4 = 272, then the quadratic equation
whose roots are a and b can be
(b) x2 – 2x + 12 = 0
(a) x2 – 2x – 16 = 0
2
(d) none of these
(c) x – 2x – 8 = 0
Ans. (c)
Solution: a + b = 2 fi a2 + b2 + 2ab = 4
fi a2 + b2 = 4 – 2ab fi (a 2 + b 2)2 = 16 – 16ab + 4a 2b 2
fi a 4 + b 4 + 2a2b 2 = 16 – 16ab + 4a 2b 2
fi 272 + 2(ab)2 = 16 – 16ab + 4(ab)2
fi 2(ab)2 – 16(ab) – 256 = 0
fi (ab)2 – 8(ab) – 128 = 0
fi (ab – 16) (ab + 8) = 0
fi ab = 16 or ab = – 8
Thus, required equation is either
x2 – 2x + 16 = 0
2
or
x – 2x – 8 = 0
\ Required answer is (c).
Example 74: If a π b and a 2 = 5a – 3, b 2 = 5b – 3, then
a
b
and
is
the equation whose roots are
b
a
(a) 3x 2 – 25x + 3 = 0 (b) x 2 – 5x + 3 = 0
(d) 3x 2 – 19x + 3 = 0
(c) x 2 + 5x – 3 = 0
Ans. (d)
Solution: a, b are roots of x2 = 5x – 3 or x2 – 5x + 3 =
0. Thus,
a + b = 5, ab = 3.
19
x + 1 = 0 or 3x 2 – 19x + 3 = 0.
3
Example 75: If a π b and difference between the roots
of the equations x2 + ax + b = 0 and x2 + bx + a = 0 is the
same, then
(a) a + b + 4 = 0
(b) a + b – 4 = 0
(c) a – b + 4 = 0
(d) a – b – 4 = 0
Ans. (a)
Solution: Let a, b be the roots of x2 + ax + b = 0 and g,
d be the roots of x2 + bx + a = 0.
We are given
|a – b | = |g – d |
fi
|a – b |2 = |g – d |2
fi
(a + b)2 – 4ab = (g + d)2 – 4gd
fi
a2 – 4b = b2 – 4a
fi
a2 – b2 + 4(a – b) = 0
fi
(a – b)(a + b + 4) = 0
As
a π b, a + b + 4 = 0.
Example 76: Product of real roots of the equation
t2x2 + |x| + 9 = 0. t ΠR is always
(a) positive
(b) negative
(c) zero
(d) does not exist.
Ans. (d)
Solution: Note that t2x2 + |x| + 9 ≥ 0 + 9 > 9 ⁄ x Œ R,
Thus, t2x2 + |x| + 9 = 0 does not have real roots.
Example 77: Let f(x) =
of f(x) is
(a) [0, •)
(c) [3, •)
Ans. (b)
Solution: Let y =
each x ΠR.
fi
x2 - 2 x + 4
x2 + 2 x + 4
, x ΠR. The range
(b) [1/3, 3]
(d) [0, 3]
x2 - 2 x + 4
x2 + 2 x + 4
=
( x - 1)2 + 3
> 0 for
( x + 1)2 + 3
(y – 1)x2 + 2(y + 1)x + 4(y – 1) = 0
As x is real,
4(y + 1)2 – 16(y – 1)2 ≥ 0
Quadratic Equations 3.21
fi (y + 1 + 2y – 2) (y + 1 – 2y + 2) ≥ 0
fi 3(y – 1/3) (y – 3) £ 0
fi 1/3 £ y £ 3.
Example 78: The values of a ΠR for which one root of
the equation f(x) = x2 – (a + 1)x + a2 + a – 8 = 0 exceeds 2
and the other is less than 2 are given by
(a) 3 < a < 10
(b) a > 0
(c) – 2 < a < 3
(d) a £ – 2
Ans. (c)
TIP: If f(x) = ax2 + bx + c, a, b, c ΠR with a > 0 and f(a )
< 0 for some a ΠR then D = b2 Р4ac > 0, and hence f(x) = 0
has real and distinct roots.
Solution: As one of the roots is less than 2 and other is
more than 2, f(2) < 0.
fi
22 – (a + 1) (2) + a2 + a – 8 < 0
a2 – a – 6 < 0 fi (a + 2) (a – 3) < 0
– 2 < a < 3.
fi
fi
Example 79: If one root of the equation x2 + px + 12 = 0
is 4, while equation x2 + px + q = 0 has equal roots, then the
value of q is
(a) 3
(b) 12
49
(c)
(d) 4
4
Ans. (c)
Solution: 42 + 4p + 12 = 0 fi p = – 7
Since x2 – 7x + q = 0 has equal roots, we get 49 – 4q = 0
49
fiq=
4
Example 80: Suppose a is an integer and x1 and x2 are
positive real roots of x2 + (2a – 1) x + a2 = 0, then value of
x1 - x2 is
(a) 1
(b) 2
(c) a
(d) 1 – 4a
Ans.(a)
Solution: Two real roots are positive implies 2a – 1 £ 0,
that is, a £ 1/2. As a is an integer, we get a £ 0. Now,
x1 - x2
2
=
(
x1 - x2
2
)
= x1 + x2 – 2 x1 x2
= (1 – 2a) – 2 a 2 = 1 – 2a – 2 |a|
= 1 – 2(a + |a|) = 1
fi
x1 - x2
=1
Example 81: If a, b are two real number satisfying the
relations 2a2 – 3a – 1 = 0 and b2 + 3b – 2 = 0 and ab π 1, then
ab + a + 1
is
value of
b
(a) – 1
(c) 1
Ans. (c)
(b) 0
(d) 2
3 2
Solution: b2 + 3b – 2 = 0 fi 1 + - 2 = 0
b b
2 3
fi
=
0
1
b2 b
Thus, a and 1/b are roots of 2x2 – 3x – 1 = 0
3
a -1
1
\
a+ =
and =
2
b 2
b
ab + a + 1
1 a
3 1
Now,
= a + + = - =1
b
b b
2 2
Example 82: If a, b are roots of x2 – 2x – 1 = 0, then
value of 5a4 + 12b3 is
(a) 153
(b) 169
(c) 183
(d) none of these
Ans. (b)
Solution: We have a 2 = 2a + 1
fi
a4 = (2a + 1)2 = 4a 2 + 4a + 1
= 4(2a + 1) + 4a + 1 = 12a + 5, and
b2 = 2b + 1 fi
Thus,
b3
= 2b2 + b = 2(2b + 1) + b
= 5b + 2
5a4 + 12b3= (60a + 25) + (60b + 24)
= 60(a + b ) + 49 = 60 (2) + 44 = 169
Example 83: Suppose a, b, c are the lengths of three
sides of a DABC, a > b > c, 2b = a + c and b is a positive
integer. If a2 + b2 + c2 = 84, then value of b is
(a) 7
(b) 6
(c) 5
(d) 4
Ans. (c)
Solution: We have
1
1
ac =
[(a + c)2 – (a2 + c2)] =
[4b2 – (84 – b2)]
2
2
= 5b2/2 – 42
Thus, a and c are the roots of the equation
x2 – 2bx + (5b2/2 – 42) = 0
As a and c are distinct real numbers the discriminant of the
above must be positive, that is, 4b2 – 4(5b2/2 – 42) > 0
fi
6b2 < 168 or b2 < 28.
Also,
ac > 0 fi 5b2 > 84.
\
84/5 < b2 < 28.
As b is a positive integer, we get b = 5.
Example 84: Suppose a, b Œ R, a π 0 and 2a + b π 0.
A root of the equation
(a + b) (ax + b) (a – bx) = (a2x – b2) (a + bx) is
a + 2b
2a + b
(a)
(b)
2a + b
a + 2b
3.22
Complete Mathematics—JEE Main
(c) –
a - 2b
a + 2b
(d) -
Solution: As roots of x2 – ax + b = 0 are real and distinct, a2 – 4b > 0 fi 4b < a2.
a + 2b
2a + b
Ans. (d)
Solution: We can write the quation as
2
[a b + (a + b) ab]x2 + [(a3 – b3) – (a + b) (a2 – b2)]x
– ab2 – ab(a + b) = 0
Thus, a2 – 1 £ 4b < a2.
fi (2a + b)x2 – (a – b)x – (a + 2b) = 0
Since the sum of coefficients is 0 one of the roots is 1 and the
a + 2b
other root is –
.
2a + b
Example 85: Let p and q be real numbers such that p
π 0, p3 π q and p3 π – q. If a and b are nonzero complex
numbers satisfying a + b = – p and a3 + b3 = q, then a
a
and b as its roots is
quadratic equation having
b
a
3
2
3
(a) (p + q)x – (p + 2q)x + (p3 + q) = 0
(b) (p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0
(c) (p3 – q)x2 – (5p3 – 2q)x + (p3 – q) = 0
(d) (p3 – q)x2 – (5p3 + 2q)x + (p3 – q) = 0
Ans. (b)
Solution: q = a3 + b3 = (a + b)3 – 3ab (a + b)
= – p3 + 3abp
p3 + q
fi
ab =
3p
We have
a b
a2 + b2
+
=
b a
ab
2
=
=
=
and
If a, b are roots of x2 – ax + b = 0, then
|a – b | £ 1 fi (a – b)2 £ 1
fi
(a + b)2 – 4ab £ 1
fi
a 2 – 4b £ 1 fi a2 – 1 £ 4b.
Example 87: If a is the minimum root of the equation
x2 – 3|x| – 2 = 0, then value of – 1/a is
(a)
p
3 p3 - 2 p3 - 2q
p3 + q
|x| = 1 3 ± 17
2
|x| ≥ 0, we get |x| = 1 3 + 17
2
1
x=±
3 + 17
2
(
fi
or
)
(
As
(
fi
\
a=–
)
)
1
2
17 - 3
1
=
3 + 17 fi - =
a
4
17 + 3
2
(
)
Example 88: Let f(x) =
The range of f is
x2 - 2 x + 3
x2 - 2 x - 8
, x ΠR Р{Р2, 4}
Ê -2 ˘
(a) Á , 1˙
Ë 9 ˚
-2
(b) R – Ê , 1ˆ
Ë 9 ¯
-2 ˘
Ê
(c) Á - •, ˙
Ë
9˚
Ê -2 ˘
(d) R – Á , 1˙
Ë 9 ˚
=
p3 - 2q
Solution: Let y =
p3 + q
( p3 - 2 q )
x+1=0
p3 + q
(p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0
Example 86: Suppose a, b ΠR. If the roots of x2 Рax
+ b = 0 are real and distinct and differ by at most 1, then 4b
lies in the interval
(b) [a2 – 1, a2)
(a) (a2 – 1, a2)
2
(d) (a2, •)
(c) (a – 1, •)
Ans. (b)
x2 - 2 x + 3
x2 - 2 x - 8
, x ΠR Р{Р2, 4}
Note that y π 1.
Now,
(y – 1)x2 – 2x(y – 1) – (8y + 3) = 0
Thus, required quadratic equation is
x –
(d) – 3
Ans. (d)
a b
◊
=1
b a
2
)
17 + 3 4
=0
2
-2
(
(b)
Solution: x2 – 3|x| – 2 = 0 can be written as |x|2 – 3|x| – 2
2
( p + q )/ 3 p
)
17 - 3 4
(c) 2
Ans. (a)
(a + b ) - 2ab
(a + b )
=
-2
ab
ab
3
(
As x is real
fi
fi
fi
4(y – 1)2 + 4(y – 1) (8y + 3) ≥ 0
(y – 1) [y – 1 + 8y + 3] ≥ 0.
(y – 1) (y + 2/9) ≥ 0
y £ – 2/9 or y > 1. [ y π 1]
Thus,
y ΠR Р(Р2/9, 1].
Example 89: Suppose a, b, c are three non-zero real
numbers. The equation
Quadratic Equations 3.23
x2 + (a + b + c)x + (a2 + b2 + c2) = 0
(1)
Example 93: Let for a, a1 π 0, a π a1
f(x) = ax2 + bx + c, g(x) = a1x2 + b1x + c1
has
(a)
(b)
(c)
(d)
Ans. (d)
two negative real roots,
two positive real roots,
two real roots with opposite signs,
no real roots
and
p(–2) = 2, then the value of p(2) is
(a) 3
(b) 9
(c) 6
(d) 18
Ans. (d)
= – {(b2 + c2 – 2bc) + (c2 + a2 – 2ca) + (a2 + b2 –
2ab) + (a2 + b2 + c2)}
= – [(b – c)2 + (c – a)2 + (a – b)2 + (a2 + b2 + c2)] < 0
Thus, (1) cannot have real roots.
Example 90: Suppose a, b ΠR. If the equation
x2 – (2a + b)x + (2a2 + b2 – b + 1/2) = 0
has two real roots, then
1
1
(a) a = , b = – 1
(b) a = – , b = 1
2
2
(c) a = 2, b = 1
(d) a = – 2, b = – 1
Ans. (a)
(1)
Solution: As (1) has real roots,
(2a + b)2 – 4 (2a2 + b2 – b + 1/2) ≥ 0
fi
4a2 + 3b2 – 4ab – 4b + 2 £ 0
fi
(2a – b)2 + 2(b – 1)2 £ 0 fi b = 1, a =
1
2
Example 91: The equation esin x – e– sin x = 4 has:
(a) no real roots
(b) exactly one real root
(c) exactly four real roots
(d) infinite number of real roots.
Ans. (a)
Solution: Put esin x = y. Note that 1 e £ y £ e.
Also, the given equation can be written as
fi
or
y2 – 4y – 1 = 0
y = 2 ± 5.
As 1 e £ y £ e, none of the two values of y is possible.
2
4
If p(x) = 0 only for x = –1
and
Solution: Discriminant D of (1) is given by
D = (a + b + c)2 – 4(a2 + b2 + c2)
y– 1 y =4
p(x) = f(x) – g(x).
Solution: As a π a1, p(x) = f(x) – g(x) is a quadratic
polynomial.
As
p(x) = 0 only for x = –1, we get
p(x) = k(x +1)2 where k = a –a1
As
Thus,
Example 94: If a, b, c are non-zero rational numbers
such that a + b + c = 0, then the roots of the equation
(b + c – a) x2 + (c + a – b) x + (a + b – c) = 0
are
(a) both rational
(b) both irrational
(c) both purely imaginary
(d) one rational and one irrational
Ans. (a)
Solution: As a + b + c = 0, we can write the equation
2ax2 + 2bx + 2c = 0
or
ax2 + bx + c = 0.
Note that x = 1 satisfies this equation and its other root is c/a.
Example 95: If a and b are the roots of the equation
ax2 + bx + c = 0, then roots of
ax2 – bx(x – 1) + c(x – 1)2 = 0
(1)
are
1 1
,
(b) 1 , b
(a)
a b
a
1
a
b
,
(c)
(d) a ,
1+a 1+ b
b
Ans. (c)
Solution: We can write (1) as
2
Example 92: If x – 3x + 2 is a factor of x – ax + b then
the equation whose roots are a and b is
(b) x2 – 9x + 20 = 0
(a) x2 – 9x – 20 = 0
2
(c) x2 + 9x + 20 = 0.
(c) x + 9x – 20 = 0
Ans. (b)
Solution: As x2 – 3x + 2 = (x – 1) (x – 2) is a factor of x4
– ax2 + b, x = 1 and x = 2 are zeros of x4 – ax2 + b, therefore,
1 – a + b = 0, 16 – 4a + b = 0
fi
a = 5, b = 4.
Thus, equation whose roots a and b is x2 – 9x + 20 = 0.
p(–2) = 2, we get 2 = k(–2 + 1)2 = k.
p(2) = 2(2 + 1)2 = 18
x ˆ2
Ê
Ê -x ˆ
aÁ+ bÁ
+c=0
˜
Ë x - 1¯
Ë x - 1˜¯
fi
-x
= a, b
x -1
fi
x=
a
b
,
1+a 1+ b
Example 96: If 3p2 = 5p + 2 and 3q2 = 5q + 2 then the
equation whose roots 3p – 2q and 3q – 2p is
(a) x2 – 5x + 100 = 0
(c) 3x2 + 5x + 100 = 0
Ans. (b)
(b) 3x2 – 5x – 100 = 0
(d) 5x2 – x + 7 = 0
3.24
Complete Mathematics—JEE Main
Solution: Note that p, q are roots of
3x2 = 5x + 2 or 3x2 – 5x – 2 = 0
p + q = 5/3, pq = –2/3.
a = 3p – 2q, b = 3q – 2p,
Let
then
and
a + b = p + q = 5/3
a b = (3p – 2q) (3q – 2p)
˘
È
1
1
1
1 ...., ˙
y = 6 + log3/2 Í
444Í3 2
˙
3 2
3 2
3 2
Î
˚
is
(a) 2
(b) 3
(c) 4
(d) 8
Ans. (c)
= – 6(p2 + q2) + 13pq
= – 6(p + q)2 + 25pq = –100/3.
Thus, required quadratic equation is
Solution: Let x =
1
3 2
4-
1
3 2
4-
2
x – (5/3)x – 100/3 = 0
fi
2
3x – 5x – 100 = 0
Example 97: The number of distinct real roots of the
equation
(x + 3)4 + (x + 5)4 = 16
(1)
is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (b)
Solution: Put x + 4 = t, so that (1) becomes
(t – 1)4 + (t + 1)4 = 16
fi
2(t4 + 6t2 + 1) = 16
fi
t4 + 6t2 – 7 = 0
fi
t2 = 1, –7 fi t = ±1, 7i
Thus, the equation (1) has two real roots.
Example 98: If a, b, c, ΠR and the equation x2 + (a + b)x
+ c = 0 has no real roots, then c(a + b + c) more than
(a) 2
(b) – 2
(c) 0
(d) none of these.
Ans. (c)
2
Solution: We have f(x) = x + (a + b) x + c > 0 " x ΠR
Thus, f(0) > 0, f(1) > 0
fi c(a + b + c) > 0
Example 99: The value of
then
fi
fi
fi
As
Thus,
1
3 2
4-
1 ....,
3 2
1
4-x
3 2
18x2 + x – 4 = 0
(9x – 4) (2x + 1) = 0
x = 4/9, –1/2
x > 0, we get x = 4/9.
y = 6 + log3/2 (4/9)
= 6 + log3/2 (3/2)–2 = 4
x=
Example 100: The sum of the roots of the equation
x + 1 – 2 log2 (2x + 3) + 2 log4 (10 – 2–x) = 0 (1)
is
(a) log211
(c) log213
Ans. (a)
fi
fi
fi
fi
(b) log212
(d) log214
Solution: Put 2x = t, then (1) can be written as
2
1ˆ
Ê
1 + log2t – 2 log2 (t + 3 ) + log2 ÁË 10 - ˜¯ = 0
t
2
1 + log2t – log2 (t + 3)2 + log2(10t – 1) – log2t = 0
log2(10t – 1) / (t + 3)2 = –1
1
10t – 1 = (t + 3)2
2
t2 – 14t + 11 = 0
(2)
If t1 = 2 x 1 , t2 = 2 x2 are roots of (2),
then,
t1t2 = 2 x1 . 2 x2 = 11
fi x1 + x2 = log211
Assertion-Reason Type Questions
Example 101: Suppose a, b, c, p Œ R a π 0, c π 0 and
b2 – 4ac > 0.
Statement-1: If the Roots of f(x) = ax2 + bx + c = 0 are
symmetrically placed on the real line about the point p,
then p = b/2a and a(ap2 + bp + c) < 0
Statement-2: If the roots of ax2 + bx + c = 0 are equal in
magnitude but opposite in signs, then b = 0 and ac < 0.
Ans. (a)
Solution: If a, – a are roots of ax2 + bx + c = 0,
then
0 = a + (– a) = –b/a fi b = 0
Also, a(– a) = c/a.
As c π 0, – a π 0 and therefore c/a = – a2 < 0 fi ac < 0.
Thus, Statement-2 is true.
Quadratic Equations 3.25
If p – a and p + a are roots of f(x) = ax2 + bx + c = 0, then
a, – a are roots of f(x + p) = a(x + p)2 + b(x + p) + c = 0,
that is, a, – a are roots of
ax2 + (2ap + b)x + ap2 + bp + c = 0.
By Statement-2: 2ap + b = 0 fi p = –b/2a and a(ap2 + bp +
c) < 0.
Example 102: Suppose a, b, c Œ R, a π 0, b2 – 4ac > 0.
Statement-1: If a, b are roots of f(x) = ax2 + bx + c = 0, such
that a < – 1 and b > 1, then a + |b| + c < 0.
Statement-2: If a, b are roots of f(x) = ax2 + bx + c = 0 such
that a < – 1, b > 1, then ac < 0.
Ans. (d)
Solution: If a < –1, b > 1, then a < 0, b > 0
c
< 0 ¤ ac < 0.
fi
ab < 0 ¤
a
Thus, Statement-2 is true.
b
c
As ax2 + bx + c = 0 and x2 + x +
= 0 have the same
a
a
roots, roots of
x2 +
b
c
x+
= 0 are a, b.
a
a
b
c
x + = (x – a) (x – b).
a
a
Note that g(x) < 0 for a < x < and a < – 1 < 1 < b
\
g(– 1) < 0, g(1) < 0
\
g(x) = x2 +
fi
b c
b c
1 – + < 0, 1 + + < 0
a a
a a
fi
1+
b
c
+ < 0.
a
a
We cannot conclude a + |b| + c < 0 unless a > 0.
Example 103: Suppose a, b, c ΠR
Statement-1: If a > 0, b2 – 4ac < 0, and f(x) = a + bx + cx 2
then domain of f is R
Statement-2: ax2 + bx + c > 0 " x ΠR if a > 0 and b2 Р4ac
> 0.
Ans. (a)
Solution: Statement-2 is true. See Theory.
As ax2 + bx + c > 0 " x ΠR, we get c = a(0)2 + b(0) + c > 0.
Thus, cx2 + bx + a > 0 " x ΠR.
Therefore, domain of f is R.
Hence, Statement-1 is true and
Statement-2 is a correct explanation for it.
Example 104: Suppose a, b, c, d ΠR.
Statement-1: If a < b < c < d, then the quadratic equation
(x – a) (x – c) + 2(x – b) (x – d) = 0 has real and distinct roots.
Statement-2: Let f(x) = ax2 + bx + c:
If f(a) f(b ) < 0 for some a, b ΠR, then f(x) = 0 has real and
distinct roots.
Ans. (a)
Solution: As f(x) = ax2 + bx + c is a continuous function
and f(a) f(b) < 0, there is at least one g ΠR lying between
a and b such that f(g ) = 0. As a, b, c ΠR and one root of
f(x) = 0 is real, other must be real. Also, roots are distinct.
Therefore, Statement-2 is true.
Now, let f(x) = (x – a) (x – c) + 2 (x – b) (x – d),
then f(a) = 2(a – b) (a – d) > 0,
f(b) = (b – a) (b – c) > 0
Thus, f(x) = 0 has a real root lying in the interval (a, b). As
coefficients are real and one of the roots is real, the quadratic
equation must have real and distinct roots.
Example 105: Suppose a, b, c a ΠR
Statement-1: The quadratic equation
(x – sin a) (x – cos a) + 3 = 0 has imaginary roots.
Statement-2: If a > 0 and b2 – 4ac < 0, then the graph of
y = ax2 + bx + c lies above the x-axis.
Ans. (a)
Solution: For truth of Statement-2, see Theory.
We can write
(x – sin a) (x – cos a) – 3 = 0
as
f(x) = 0
where
f(x) = x2 – (sin a + cos a) + sin a cos a +
3
coefficient of x2 = 1 > 0, and
b2 – 4ac = (sin a + cos a)2 – 4(sin a cos a +
= (sin a – cos a)2 – 4 3
3)
2
£ ( 2) – 4 3 < 0
as maximum possible value of sin a – cos a is 2 .
Thus graph of y = f(x) lies above the x-axis and hence
f(x) = 0 has imaginary roots.
Example 106: Suppose a, b, c Œ R and a π 0.
Statement-1 If ac > 0, ab < 0, b2 – 4ac > 0 the equation
ax4 + bx2 + c = 0 has four distinct roots.
Statement-2 If ac > 0, ab < 0, then the equation ax2 + bx +
c = 0 has distinct positive roots.
Ans. (c)
Solution: Statement-2 is false as x2 – x + 1 = 0 satisfies
the condition ac > 0 and ab < 0 but x2 – x + 1 = 0 does not
have positive roots.
Statement-1 is true. Put x2 = y, so that ay2 + by + c = 0.
As b2 – 4ac > 0. This equation has two distinct roots.
b
ab
c ac
= - 2 > 0 and = 2 > 0 both the roots of ay2 + by
a
a a
a
+ c = 0 are positive. If the roots are a, b > 0, then x2 = a, b
As -
fi x=±
a, ± b .
3.26
Complete Mathematics—JEE Main
Thus, ax4 + bx2 + c = 0 has four distinct roots.
Example 107: Suppose a, b, c Œ R, a π 0, and let f(x)
= ax2 + bx + c.
Statement-1: If f(x) = 0 has imaginary roots and a + c > 0,
then f(x) > 0 " x ΠR.
Statement-2: If f(x) > 0 " x Œ R, then g(x) = f(x) + f ¢(x) +
f ¢¢(x) > 0 " x Œ R.
Ans. (b)
¤
Solution: f(x) > 0 " x ΠR,
a > 0, b2 – 4ac < 0.
We have g(x) = ax2 + bx + c + (2ax + b) + 2a
= ax2 + (b + 2a)x + c + b + 2a
Note that
(b + 2a)2 – 4a (c + b + 2a)
= (b2 – 4ac) – 4a2 < 0
and
a > 0, thus g(x) > 0 " x ΠR
\ Statement-2 is true.
As f(x) = 0 has imaginary roots, f(x) > 0 " ΠR or f(x) < 0
"xŒR
We have f(1) + f(–1) = 2(a + c) > 0,
therefore f(x) > 0 " x ΠR.
Thus, statement-1 is true but statement-2 is not a correct
explanation for it.
Example 109: Suppose a, b, c Œ I, a π 0 and f(x) = ax2
+ bx + c.
Statement-1: If f(x) = 0 has no rational roots, then
1
Ê pˆ
f Á ˜ ≥ 2 " p, q Œ I, q π 0 and
Ë q¯ q
Statement-2: If a +
f(x), then its other zero must be a –
( x - b) ( x - c)
( x - c) ( x - a)
+b
(a - b) (a - c )
(b - c ) (b - a )
+ c
( x - a ) ( x - b)
= x " x ΠR.
(c - a ) (c - b )
Statement-2: If Ax2 + Bx + C = 0 for x = a, b, c, then A =
B = C = 0.
Ans. (a)
Solution: If three distinct values of x satisfy a quadratic
equation, then it is identically equal to zero. Thus, Statement-2 is true.
Let
f(x) = a
Let
Solution: Statement-2 is true. (See Theory)
x = p/q, p, q Œ I and q π 0.
Then
Ê pˆ
1
f Á ˜ = 2 (ap2 + bpq + cq2)
Ë q¯
q
f(x)= 0 has no rational roots, f(p/q) π 0
fi ap2 + bpq + cq2 π 0
As
As ap2 + bpq + cq2 is an integer, we get |ap2 + bpq + cq2| ≥ 1.
1
Ê pˆ
Ê bˆ
Thus, q2 f Á ˜ ≥ 1 fi f Á ˜ ≥ 2 .
Ë q¯ q
Ë q¯
Example 110: Suppose a, b, c ΠR and a < b < c.
Let f : R Æ R be defined by
Ï ( x - a )( x - c) if x π b
Ô
x-b
f ( x) = Ì
ÔÓ
0
if x = b
Statement-1: Range of f is R.
Statemetnt-2: For each y ΠR, y = f(x) has exactly two
distinct solutions.
Ans. (c)
Solution: Statement-2 is false as f(a) = f(b) = f(c) = 0
fi
f(x) = 0 has three distinct solutions.
Note that Statement-1, put t = x – b, so that
f(x) =
( x - a ) ( x - b)
– x.
(c - a ) (c - b )
Then f(x) = 0 is a quadratic and f(a) = f(b) = f(c) = 0.
Thus, f(x) ∫ 0.
Therefore, Statement-1 is true and statement-2 is a correct
reason for it.
(t + b - a ) (t + b - c )
t
= t – (a + c – 2b) +
( x - b) ( x - c)
( x - c) ( x - a)
+b
(a - b) (a - c )
(b - c ) (b - a )
+ c
b.
Ans. (b)
Example 108: Suppose a, b, c ΠR are three distinct real
numbers.
Statement-1:
a
b , a, b Œ Q and b π 0 is a zero of
= x – (a + c – b) +
( a - b ) (c - b )
t
( a - b ) (c - b )
x-b
note that f is continuous on (– •, b). Also, lim f(x) = – •
xÆ - •
and lim f(x) = •
xÆ b-
Thus, range of f is (–•, •) = R.
\ Statement-1 is true.
Quadratic Equations 3.27
LEVEL 2
Straight Objective Type Questions
Example 111: Suppose a, b, c, d ΠR and 2ac = b + d.
Consider the quadratic equations x2 + 2ax + b = 0 and x2 +
2cx + d = 0. Then
(a) none of these equations have real roots
(b) both the equations have real roots
(c) exactly one of the equation has real roots
(d) at least one of the equation has real roots
Ans. (d)
Solution: Let D1 be the discriminant of x2 + 2ax + b =
0 and D2 be the discriminant of x2 + 2cx + d = 0
Then
D1 + D2 = 4 (a2 – b) + 4(c2 – d)
= 4(a2 + c2) – 8ac
= 4(a – c)2 ≥ 0
fi at least one of D1, D2 is non-negative. Thus, at least one
of the equations has real roots.
Example 112: If the roots of the equation
1
1
1
(1)
+
=
x+a x+b c
are equal in magnitude but opposite in sign, then their product is
1 2
1 2
(a + b2)
(b) –
(a + b2)
(a)
2
2
1
1
(c)
ab
(d) –
ab
2
2
Ans. (b)
Solution: The equation (1) can be written as
c(x + b + x + a) = (x + a) (x + b)
(2)
or
x2 + (a + b – 2c)x + ab – ac – bc = 0
Let a and – a be the roots of (2) then
1
(a + b)
0 = a + (– a) = a + b – 2c fi c =
2
Also,
2a ( – a) = 2ab – 2 (a + b)c = 2ab – (a + b)2
= – (a2+ b2)
1
fi
a (– a) = - (a 2 + b2 )
2
Example 113: If c, d are roots of x2 – 10ax – 11b = 0
and a, b are root of x2 – 10cx – 11d = 0, then value of a + b
+ c + d is
(a) 1210
(b) – 1
(c) 2530
(d) – 11
Ans. (a)
Solution: c + d = 10a
(1)
a + b = 10c
(2)
Subtracting (1) from (2) we get
(a – c) + (b – d) = 10 (c – a)
fi
b – d = 11(c – a)
(3)
As c is a root of x2 – 10ax – 11b = 0,
we get
c2 – 10ac – 11b = 0
(3)
2
(4)
Similarly,
a – 10ac – 11d = 0
Subtracting (4) from (3), we get
c2 – a2 = 11(b – d)
fi (c – a) (c + a) = (11)11(c – a)
fi
c + a = 121
\
a + b + c + d = 10(a + c)
[from (1) and (2)]
= 10(121) = 1210
Example 114: If a, b are the roots of the equation ax2 +
bx + c = 0, then the value of a3 + b3 is
(a)
(c)
Ans. (c)
3abc + b3
a3
3abc - b3
a
3
(b)
(d)
a 3 + b3
3abc
(
- 3abc + b3
a
)
3
Solution: We have
-b
c
a+b=
, ab =
a
a
Now,
a3 + b3 = (a + b)3 – 3ab(a + b)
3
3c Ê - b ˆ 3abc - b3
Ê - bˆ
= Á ˜ - Á ˜=
Ë a ¯
aË a ¯
a3
Example 115: Suppose a ΠI and the equation
(x – a) (x – 5) = 3 has integral roots, then the set of
values which a can take is:
(a) f
(b) {– 11, – 13}
(c) {3, 7}
(d) {– 3, – 7}
Ans. (c)
Solution: Let m ΠI be a roots of (x Рa) (x Р5) = 3,
then
(m – a) (m – 5) = 3
As m – a and m – 5 are integers, m – 5 = ± 1 or m – 5 = ± 3
fi m = 2, 4, 6, 8.
Thus, a = 3, 7.
3.28
Complete Mathematics—JEE Main
Example 116: If the ratio of the roots of the equation
x + bx + c = 0 is the same as that of the ratio of the roots of
x2 + qx + r = 0, then
(b) cq2 = rb2
(a) br2 = qc2
2 2
2 2
(c) q c = b r
(d) bq = rc
Ans. (b)
2
Solution: Let a, b be the roots of the equation x2 + bx
+ c = 0 and g, d be the roots of the equation x2 + qx + r = 0.
We are given
a g
a - b g -d
=
= fi
b d
a + b g +d
2
2
fi
Êa - bˆ
Êg -dˆ
ÁË a + b ˜¯ = ÁË g + d ˜¯
fi
(a + b )2 - 4ab (g + d )2 - 4gd
=
(a + b )2
(g + d )2
fi
fi
ab
2
(a + b )
c
2
(- b)
=
=
gd
But
f ¢(x) = 3ax2 + 2bx + c
2
Hence, 3ax + 2bx + c = 0 has a root in [0, 1].
Example 119: If a < b < c < d, then the equation
3(x – a) (x – c) + 5(x – b) (x – d) = 0
has
(a) real and distinct roots
(b) real and equal roots
(c) purely imaginary roots
(d) none of these
Ans. (a)
Solution: Let f(x) = 3(x – a) (x – c) + 5(x – b) (x – d)
Since f is a polynomial, f is continuous on R. Also, since
a < b < c < d,
f(a) = 5(a – b) (a – d) > 0
f(b) = 3(b – a) (b – c) < 0
f(c) = 5(c – b) (c – d) < 0
f(d) = 3(d – a) (d – c) > 0
(g + d )2
r
2
(- q )
fi cq2 = rb2
Example 117: If a and b are the non-zero distinct roots
of x2 + ax + b = 0, then the least value of x2 + ax + b is
(a) 2/3
(b) 9/4
(c) – 9/4
(d) 1
Ans. (c)
Solution: We have a + b = – a, ab = b
As
b π 0, we get a = 1
\
1+b=–1fib=–2
Thus,
x2 + ax + b = x2 + x – 2
Ê
= Áx +
Ë
\
1ˆ 2 9
9
˜¯ - ≥ 2
4
4
least value of x2 + ax + b is – 9/4 which is attained at
x = – 1/2.
Example 118: If a + b + c = 0, then the quadratic
equation 3ax2 + 2bx + c = 0 has
(a) at least one root in [0, 1]
(b) one root in [2, 3] and other is [ – 2, – 1]
(c) imaginary roots
(d) none of these
Ans. (a)
Solution: Let f(x) = ax3 + bx2 + cx, x Π[0, 1]. Since f is
a polynomial function, f is differentiable on the whole real
line and in particular on [0, 1].
Also,
f(0) = 0 and f(1) = a + b + c = 0.
Thus, all the conditions in the hypothesis of the Rolle’s
theorem are satisfied. By the Rolle’s theorem there exists at
least one a Π(0, 1) such that
f ¢(a) = 0
Fig. 3.18
As f is continuous on R, y = f(x) crosses x-axis at least
once between a and b and once between c and d. See
Fig. 3.18.
Thus, f has two distinct real roots, one lying between a and
b and one lying between c and d.
( x - a) ( x - c)
Example 120: For real x, the function
x-b
will assume all real values provided
(a) a < b < c
(b) b < c < a
(c) c < a < b
(d) none of these
Ans. (a)
x 2 - ( a + c ) x + ac
x-b
x-b
2
fi
y(x – b) = x – (a + c)x + ac
2
(1)
fi
x – (a + c + y)x + ac + by = 0
Since x is real, the discriminant of (1)
(a + c + y)2 – 4(ac + by) ≥ 0
fi
y2 + 2(a + c)y + (a + c)2 – 4ac – 4by ≥ 0
2
(2)
fi
y + 2(a + c – 2b)y + (a – c)2 ≥ 0
Solution: Let y =
( x - a) ( x - c)
=
Since y takes all real values, (2) is possible if and only if
4(a + c – 2b)2 – 4(a – c)2 < 0
[∵ coeff. of y2 = 1> 0]
¤
(a + c – 2b + a – c) (a + c – 2b – a + c) < 0
Quadratic Equations 3.29
¤
¤
¤
(2a – 2b) (2c – 2b) < 0
(a – b) (c – b) < 0
b lies between a and c. Thus, one of the
possibilities a < b < c.
Example 121: Let a, b, c Œ R and a π 0. If a is a root of
a2 x2 + bx + c = 0, b is a root of a2 x2 – bx – c = 0 and 0 < a
< b, then the equation a2 x2 + 2bx + 2c = 0 has a root g that
always satisfies
1
1
(a + b)
(b) g = a + b
(a) g =
2
2
(c) g = a + b
(d) a < g < b
Ans. (d)
Solution: By the hypothesis,
(1)
a2 a 2 + ba + c = 0
2 2
(2)
and
a b – bb – c = 0
where 0 < a < b, and a π 0.
Let
f(x) = a2 x2 + 2bx + 2c
Since f is a polynomial function, f is continuous on R and
hence on [a, b]. Also,
f(a) = a2 a 2 + 2ba + 2c
= 2(a2a2 + ba + c) – a2 a 2
= 2(0) – a2 a2 = – a2 a 2 < 0
2
[using (1)]
2
f ( b ) = a b + 2bb + 2c
= 3a2 b 2 – 2(a2 b 2 – bb – c)
= 3 a2 b 2 – 2(0) = 3a2 b 2 > 0 [using (2)]
Since f is continuous on [a, b] and f(a) < 0, f(b) > 0, f
must vanish at least once in (a, b).
and
Example 122: Suppose p, q, r, s ΠR and a, b be the
roots of x2 + px + q = 0 and a 4, b 4 be the roots of x2 – rx + s
= 0, then the equation x2 – 4qx + 2q2 – r = 0 has always
(a) two imaginary roots
(b) two positive roots
(c) two negative roots
(d) one positive and one negative root
Ans. (d)
Solution: We have
a + b = – p, ab = q, a 4 + b 4 = r, a 4 b 4 = s
Discriminant D of the equation
(1)
x2 – 4qx + 2q2 – r = 0
2
2
2
2
is given by D = 16q – 4(2q – r) = 8q + 4r = 4[2q + r]
= 4[2a 2 b 2 + a 4 + b 4] = 4(a 2 + b 2)2 ≥ 0
Thus, the equation (1) has real roots.
The roots of (1) are given by
4q ± D
x=
= 2ab ± (a 2 + b 2)
2
= (a + b)2, – (a – b)2
Hence, (1) has one positive and one negative root.
Example 123: The equation
x (3 / 4 )(log2 x )
2
+ log2 x - 5 / 4
=
2
(1)
has
(a)
(b)
(c)
(d)
Ans. (c)
exactly two real roots
no real root
one irrational root
none of these
Solution: Taking log of the sides in (1), we get
5˘
2
È3
ÍÎ 4 ( log2 x ) + log2 x - 4 ˙˚ log2 x = log2 ( 2 )
5ˆ
1
Ê3 2
fi
ÁË y + y - ˜¯ y = log2 2 where y = log2 x
4
4
2
fi
3y3 + 4y2 – 5y – 2 = 0
fi
3y3 – 3y2 + 7y2 – 7y + 2y – 2 = 0
fi
(y – 1) (3y2 + 7y + 2) = 0
fi
(y – 1) [3y2 + 6y + y + 2] = 0
fi
(y – 1) (y + 2) (3y + 1) = 0
fi
y = 1, – 2, – 1/3
fi
log2 x = 1, – 2, – 1/3 fi x = 2,
1 –1/3
,2
4
Thus, (1) has one irrational root viz. 2–1/3.
Example 124: Let f(x) be a quadratic expression which
is positive for all x. If g (x) = f (x) + f ¢(x) + f ¢¢(x) then for all
real x,
(a) g(x) < 0
(b) g(x) > 0
(c) g(x) = 0
(d) g(x) ≥ 0
Ans. (b)
Solution: Let f(x) = ax2 + bx + c
As f(x) > 0 " x ΠR, we must have a > 0 and b2 Р4ac < 0.
Also,
f ¢(x) = 2ax + b and f ¢¢(x) = 2a
Thus,
g(x) = ax2 + (2a + b)x + 2a + b + c.
Since a > 0 and discriminant
(2a + b)2 – 4a(2a + b + c)
= 4a2 + 4ab + b2 – 8a2 – 4ab – 4ac
[ b2 – 4ac < 0]
= – 4a2 + (b2 – 4ac) < 0
we get
g(x) > 0 " x ΠR
Example 125: If a, b are the roots of the quadratic
equation ax2 + bx + c = 0, then the quadratic equation whose
roots are a 3, b 3 is
(a) a3 y2 + (b3 – 3abc)y + c3 = 0
(b) a3 y2 + (3abc – b3)y – c3 = 0
(c) a2 y2 + 2aby + c2 = 0
(d) none of these
Ans. (a)
Solution: Put y = a3 fi a = y1/3.
As a is a root of ax2 + bx + c = 0, we get
ay2/3 + by1/3 + c = 0 fi y1/3 (ay1/3 + b) = – c
(1)
3.30
Complete Mathematics—JEE Main
Cubing both the side, we get
y(ay
1/3
3
+ b) = – c
3
3
fi
y[a y + b3 + 3aby1/3 (ay1/3 + b)] = – c3
fi
y[a3y + b3 + 3ab (– c)] = – c3 [using (1)]
fi
a3y2 + (b3 – 3abc)y + c3 = 0
Example 126: If a, b, c ΠR and the equations ax2 + bx +
c = 0 and x3 + 3x2 + 3x + 2 = 0 have two common roots, then
(a) a = b = – c
(b) a = – b = c
(c) a = b = c
(d) none of these
Ans. (c)
Solution: x3 + 3x2 + 3x + 2 = 0
fi
(x + 1)3 = – 1
fi
x + 1 = – 1, – w, – w2
fi
x = – 2, – 1, – w, – 1 – w2
As a, b, c ΠR, the roots of ax2 + bx + c = 0 are both real or
both imaginary.
\ roots of ax2 + bx + c = 0 must be – 1 – w, – 1 – w2.
b
Thus,
– 1 – w – 1 – w2 = –
a
c
and
(– 1 – w) (– 1 – w2) =
a
b
c
fi
1=
and 1 =
a
a
fi
a=b=c
\ f(x) is partly positive and partly negative on [1, 2].
fi there exist a, b Π[1, 2] such that
f(a) > 0 and f(b) < 0.
As f is continuous on [1, 2] there exists g lying between
a and b (and hence between 1 and 2) such that f(g) = 0
fi
(1 + cos8 g) (ag2 + bg + c) = 0
fi
ag2 + bg + c = 0
[ 1 + cos8 g ≥ 1]
Thus, ax2 + bx + c = 0 has at least one root in [1, 2].
Example 128: The number of real solutions of the
equation
(1)
271/x + 121/x = 2(81/x )
is
(a) 0
(b) 1
(c) infinite
(d) none of these
Ans. (a)
Solution: We can write (1) as
1/ x
1/ x
Ê 27 ˆ
Ê 12 ˆ
=2
+
ÁË ˜¯
ÁË ˜¯
8
8
or
As
and
Ê 3ˆ
ÁË ˜¯
2
3/ x
Thus,
Ê 3ˆ
ÁË ˜¯
2
3/ x
and
Example 127: Let a, b, c be non-zero real number such that
1
Ú0 (1 + cos
=
2
Ú0
8
2
x) (ax + bx + c) dx
(1 + cos8 x) (ax2 + bx + c) dx
Then the quadratic equation ax2 + bx + c = 0 has
(a) no root in (0, 2)
(b) at least one root in (1, 2)
(c) a double root (0, 2)
(d) none of these
Ans. (b)
Solution: Let f(x) = (1 + cos8 x) (ax2 + bx + c)
We are given
1
Ú0
fi
1
2
f(x) dx =
Ú f(x) dx
0
1
2
Ú0 f(x) dx = Ú0 f (x) dx + Ú1
2
f(x) dx
Ú1
If
f(x) > 0 (f(x) < 0) ⁄ x Œ [1, 2],
then
Ú1
But
Ú1
2
f(x) dx = 0
1/ x
Ê 3ˆ
+Á ˜
Ë 2¯
1/ x
< 2 if x < 0
> 2 if x > 0.
Therefore, (1) is possible if 1/x = 0.
But this is not true for any real value of x.
Example 129: If 0 < a < b < c < d, then the quadratic
equation
(1)
ax2 + {1 – a(b + c)}x + abc – d = 0
has
(a) real and distinct roots out of which one lies between c and d.
(b) real and distinct roots out of which one lies between a and b
(c) real and distinct roots out of which one lies between b and c
(d) non-real roots
Ans. (a)
f(x) dx = 0
f(x) dx > 0
Ê 3ˆ
+Á ˜
Ë 2¯
(1)
Solution: We can rewrite (1) as
ax2 – a(b + c)x + abc + x – d = 0
fi
2
3/ x
1/ x
Ê 3ˆ
Ê 3ˆ
=2
+
ÁË ˜¯
ÁË ˜¯
2
2
3/2 > 1, (3/2)t > 1 if t > 0
(3/2)t < 1 if t < 0.
(Ú
2
1
f (x) d x < 0
)
or
a(x – b) (x – c) + x – d = 0
Let
f(x) = a(x – b) (x – c) + x – d.
As a > 0, y = f(x) represents a parabola which open upwards.
See Fig. 3.19.
Quadratic Equations 3.31
b
Solution: As f(x) = ax2 + 2bx – 3c = 0 has no real roots,
f(x) > 0 " x ΠR or f(x) < 0 " x ΠR
Since
4a + 4b – 3c > 0, f(2) > 0.
\
f(x) = ax2 + 2bx – 3c > 0
fi
f(0) = – 3c > 0 fi c < 0
Also,
a > 0 and b2 + 3ca < 0
fi
c < – b2/3a fi c Œ (– •, – b2/3a)
c
d
Fig. 3.19
Also,
f(b) = b – d < 0
b
x
+ c, x ΠR Р{0}, assume all real values, then a and b satisfy
the relation
(a) ab £ 0
(b) ab ≥ 0
(c) ab ≥ 1
(d) ab £ 1
Ans. (a)
b
Solution: Let y = ax +
+ c, then
x
ax2 + (c – y)x + b = 0.
As x is real, (c – y)2 – 4ab ≥ 0
This is true for each y Œ R if ab £ 0
Example 131: Suppose a, b, c Œ R and a π b. If ax +
f(c) = c – d < 0, and f(d) = a(d – b) (d – c) > 0
Thus, f(x) = 0 has a root between – • and b and a root between c and d.
Example 130: If ax2 + 2bx – 3c = 0 has no real roots and
4
c<
(a + b), then range of c is
3
(a) (0, b)
(b) (–1, b)
(c) (– •, – b2/3a)
(d) (– •, – b/12a)
Ans. (c)
EXERCISE
Concept-based
Straight Objective Type Questions
1. The number of real solutions of x2 – 2x + 2 + |x –
1| = 0 is:
(a) 0
(b) 1
(c) 2
(d) infinite
6. The number of real solutions of x2 – 4|x| – 2 = 0
is
2. If roots of 7x2 – 11x + k = 0, k π 0 are reciprocal
of each other, then k is equal to
7. Suppose a and b are roots of the equation x2 + px
3
+ p = 0. If |a – b| = 10 , then p belongs to the
4
set
(a) {–2, 5}
(b) {–3, 2}
(c) {2, –5}
(d) {3, –5}
(a) –1
(c) 7
(b) 7/11
(d) 11/7
3. If a + b π 0 and the roots of x2 – px + q = 0 differ
by –1, then p2 – 4q equals:
(a) –1
(c) 1
(b) 0
(d) 2
2
2
4. If the equations x – ax + b = 0 and x + bx – a = 0
have a common root then
(a) a = b
(c) a – b = – 1
(b) a + b = – 1
(d) a – b = 1
2
5. If both the roots of x + x + a = 0 exceed a, then
a belongs to:
(a) (– •, –1)
(c) (0, 1)
(b) (– •, – 2)
(d) (1, •)
(a) 1
(c) 3
(b) 2
(d) 4
8. The number of solutions of
is
(a) 0
(c) 2
5+ x + x = 2
(b) 1
(d) infinite
9. Suppose a ΠR. The set of values of a for which
the quadratic equation x2 – 2(a + 1)x + a2 – 4a +
3 = 0 has two negative roots is
(a) (– •, – 1)
(b) (1, 3)
(c) (– •, 1) » (3, •) (d) f
10. Suppose a > 0, b > 0 and a + b = p/4. If tan a,
tan b are roots of x2 – ax + b = 0, then
3.32
Complete Mathematics—JEE Main
(a) a + b = 1
(c) a = b
y
(b) a + b = 1, 0 < b < 1
(d) 0 < a + b < 2
11. If b > a, and c > 0 then the equation (x – a) (x – b)
– c = 0 has:
(a)
(b)
(c)
(d)
2
12. If the quadratic equation x + 2 (k + 1)x + 9k – 5 = 0
has exactly one positive root, then k lies in the set
(a) [5/9, •)
(c) (– •, 5/9]
13. If k ΠR lies between the roots of ax + 2bx +
c = 0, then
ak2 + 2bk + c < 0
a2k2 + 2abk + ac < 0
a2k2 + 2abk + ac > 0
ak2 + 2bk + c > 0
14. If both the roots of the quadratic equation x2 – 4ax
+ 2a2 – 3a + 5 are less than 2, then a lies in the set
(a)
(b)
(c)
(d)
(9/2, •)
(– •, 9/2)
(–1, •)
(2, •)
15. Fig. 3.20 Shows graph of y = ax2 + bx + c. Then
which one of the following is not true.
(a)
(b)
(c)
(d)
0
a>0
c<0
b2 – 4ac > 0
b>0
x2
x
Fig. 3.20
16. Greatest value of the expression
(a) 2
(b) (– •, 1) » (6, •)
(d) [1, 6]
2
(a)
(b)
(c)
(d)
x1
both roots in (– •, a)
both roots in (a, b]
one root in (– •, a) and other root in (b, •)
one root in (– •, a) and other root in [a, b]
8
2
9x - 6x + 5
is
(b) 5
1
(d) 9.2
3
17. If a, b Π{1, 2, 3, 4} the number of quadratic
equation of the form ax2 + bx + c = 0 which have
non-real complex roots is:
(a) 27
(b) 35
(c) 52
(d) 56
18. The number of real roots of the equation (x – 1)2
+ (x – 2)2 + (x – 3)2 = 0 is
(a) 0
(b) 2
(c) 3
(d) infinite
(c) 8
19. Suppose 0 < a < b < c. If the roots a, b of ax2 +
bx + c = 0 are imaginary, then
c
a
(b) |b | =
(a) |a| =
a
c
(c) a + b = 0
(d) a – b = – b/2a
20. If a > 0 and both the roots of ax2 + bx + c = 0 are
more than 1, then
(a) a + b + c > 0
(b) a + b + 4c = 0
(c) a + b + c < 0
(d) a + 4b + c = 0
LEVEL 1
Straight Objective Type Questions
21. In Fig. 3.21 graph of y = ax2 + 2bx + c is given.
Which one of the following is not true?
y
22. If ax2 + 2bx + c = 0 and ax2 + 2cx + b = 0, b π
a+b+c
is equal to
c have a common root, then
a
(a) – 2
(c) 3/4
y = ax2 + bx + c
x
0
Fig. 3.21
(a) a > 0
(c) c > 0
(b) b > 0
(d) b2 < ac
(b) – 1
(d) –1/4
23. Suppose a, b, c Œ R, a π 0 and 4a – 6b + 9c < 0.
If ax2 + bx + c = 0 does not have real roots, then
b+c
is less than
a
(a) 0
(b) 1
(c) –1
(d) –2
Quadratic Equations 3.33
24. Suppose a, b are roots of 4x2 – 16x + c = 0, where
c ΠN, the set of natural numbers. The number of
values of c for which 1 < a < 2 < b < 3 is
(a) 2
(c) 4
(b) 3
(d) 9
(a) (–7, –8)
(c) (7, 8)
(b) (–8, –7)
(d) (8, 7)
a
1
1
26. Suppose a, b ΠR. If the equation =
+
x x-b x+b
is not satisfied by any real value of x, then
(b) 0 < a < 2
(d) 0 < b < a < 2
27. Suppose a Œ R, a π –1/2. Let a, b be roots of
(2a + 1)x2 – ax + a – 2 = 0 If a < 1 < b, then
(a)
1
(6 - 2 23 ) < a < 1 (b) 1 (6 - 2 23 ) < a < 1
7
7
2
1
1
< a < (6 + 2 23 ) (d) none of these
2
7
28. Suppose a, b are roots of 8x2 – 10x + 3 = 0, then
(c)
•
 (a
n
+b
n
)
is
n=0
(a)
(b)
(c)
(d)
(b) 2
(d) 0
34. If 2 + 5 i is a root of x2 – px + q = 0 where p
and q are real, then the ordered pair (p, q) is equal
to
(a) (4, 9)
(c) (3, 3)
(b) (9, 4)
(d) (2, 3)
35. If the quadratic equation 2x2 + ax + b = 0 and 2x2
+ bx + a = 0, (a π b) have a common root, the
value of a + b is
(a) – 3
(c) – 1
(b) – 2
(d) 0
36. If a, b, c, d and p are distinct real numbers such
that
(a2 + b2 + c2)p2 + 2(ab + bc + cd)p +
(b2 + c2 + d2) £ 0,
then a, b, c, d
(a) are in A.P.
(b) are in G.P.
(c) are in H.P.
(d) none of these
37. The number of real roots of the equation sin (ex) =
5x + 5–x is
30. Suppose 0 < b < c and f(x) =
(a) 2
(c) 0
(b) 1
(d) infinitely many
(b) – 2
(d) none of these
39. The roots of the equation |x2 – x – 6| = x + 2 are
(a) – 2, 1, 4
(c) 0, 1, 4
(b) 0, 2, 4
(d) – 2, 2, 4
40. The number of real roots of the equation
2
x - bc
, x ΠR,
2 x - (b + c )
then f(x) cannot lie in
(b) (– •, b)
(d) (0, b) » (b, c)
31. If a, b are the roots of x2 + px + q = 0 and g,
d are the roots of x2 + rx + s = 0, then value of
(a – g) (a – d) (b – g) (b – d) is
(a)
(b)
(c)
(d)
(b) 2
(d) 0
38. The value of a for which the equations x3 + ax + 1
= 0 and x4 + ax2 + 1 = 0 have a common root is
real and distinct
real and equal
purely imaginary
non-real complex numbers
(a) (b, c)
(c) (c, •)
(a) 4
(c) 1
(a) 0
(c) 2
7/4
3/7
6
7
29. Suppose a, b, c Œ R a π 0. If a + |b| + 2c = 0,
then roots of ax2 + bx + c = 0 are
(a)
(b)
(c)
(d)
(a) 4
(c) 1
33. The number of real solutions of x2 – 3|x| + 2 = 0 is
25. Suppose a, b Œ R, ab π 0. If all the three quadratic
equations x2 + ax + 12 = 0, x2 + bx + 15 = 0 and
x2 + (a + b)x + 36 = 0 have a common negative
root, then (a, b) is equal to
(a) 0 < a < b < 2
(c) 0 < b < 2
32. The number of real solutions of x2 + 5|x| + 4 = 0 is
(r – p)2 – (q – s)2
(r – p)2 + (q – s)2
(r – p)2 + (q – s)2 – 2rp(r – p) (q – s)
none of these
x2 – 3|x| + 2 = 0
(a) 4
(c) 2
(b) 1
(d) infinite
41. The least value of n ΠN for which (n Р4)x2 + 8x
+ n + 2 > 0 " x ΠR, is
(a) 11
(c) 8
(b) 10
(d) 7
42. Let f(x) be a quadratic expression such that f(x) <
0 " x Œ R. If g(x) = f(x) + f ¢(x) + f ¢¢(x) then for
x ΠR.
(a) g(x) < 0
(c) g(x) > 0
(b) g(x) £ 0
(d) g(x) ≥ 0
3.34
Complete Mathematics—JEE Main
43. If x + 1 is a factor of x4 + (p – 3)x3 – (3p – 5)x2
+ (2p + 9)x + 12, then value of p is
(a) – 2
(c) 1
(b) 2
(d) –1
44. If both the roots of x2 – (p – 4)x + 2e2log p – 4 = 0
are negative, then p belongs to
2 , 4)
(c) (4, •)
(a)
(–
2 , 4)
(d) none of these
(b)
(
45. Let f and g be two real valued functions and S
= {x|f(x) = 0} and T = {x|g(x) = 0}, then S « T
represent the set of roots of
(a) f(x) g(x) = 0
(b) f(x)2 + g(x)2 = 0
(c) f(x) + g(x) = 0
(d)
f (x)
=0
g (x)
46. If domain of f(x) = x 2 + bx + 4 is R, then maximum possible integral value of b is
(a) 2
(c) 4
(b) 3
(d) 5
47. If p Œ (–1, 1), then roots of the quadratic equation
(p – 1)x2 + px +
(a)
(b)
(c)
(d)
1 - p2 = 0 are
48. If a, b, c are positive real numbers, then the number
of positive real roots of the equation ax2 + bx + c
= 0 is
(b) 1
(d) infinite
2
2
49. If the roots of the equation x + p = 8x + 6p are
real, then p belongs to the interval
(a) [2, 8]
(c) [– 2, 8]
(b) [– 8, 2]
(d) [– 8, – 2]
50. If sum of the roots of the equation (a + 1)x2 + (2a
+ 3)x + (3a + 4) = 0 is – 3, then the product of
the roots is
(a) 1
(c) 2
(b) 4
(d) – 2
51. If 3 – 4i is a root of x2 – px + q = 0 where p, q
2p - q
is
ΠR, then value
p+q
(a) – 12/31
(c) – 15/31
x2 + (1 + i)x + 1 + i = 0
x2 + (1 + i)x + i = 0
x2 + 2(1 + i)x – 2 = 0
none of these
53. If a and b be the roots of the equation x2 + px –
1
= 0, where p ΠR. Then the minimum possible
2 p2
value of a2 + b 2 is
(a) 2
(c) 2 +
54. The equation
= 1 has
(a)
(b)
(c)
(d)
(b) 2 2
(d) none of these
2
x + 3- 4 x -1 + x +8 - 6 x -1
no solution
exactly one solution
exactly two solutions
more than two solution
55. The equation |x – x2 – 1| = |2x – 3 – x2| has
(a)
(b)
(c)
(d)
no solution
exactly one solution
exactly two solutions
more than two solutions
56. If sin a, cos a are the roots of the equation ax2 +
bx + c = 0, (a π 0), then
purely imaginary
non-real complex numbers
real and equal
real and distinct.
(a) 0
(c) 2
(a)
(b)
(c)
(d)
(b) – 13/31
(d) none of these
52. If x = 1 + i is a root of x3 – ix + 1 – i = 0, then
the quadratic equation whose roots are the remaining two roots of x3 – ix + 1 – i = 0 is
a2 – b2 + 2ac = 0
a2 + b2 – 2ac = 0
(a – c)2 = b2 + c2
none of these
x2 - x + 1
57. If x ΠR, and k = 2
, then
x + x +1
(a)
(b)
(c)
(d)
(a) 1/3 £ k £ 3
(c) k £ 0
(b) k ≥ 5
(d) none of these
58. If the equation x2 + bx + ca = 0 and x2 + cx + ab
= 0 have a common root and b π c, then their other
roots will satisfy the equation
(a) x2 + (b + c)x + bc = 0 (b) x2 – ax + bc = 0
(d) none of these
(c) x2 + ax + bc = 0
59. If the inequality
mx 2 + 3 x + 4
x2 + 2 x + 2
<5
is satisfied for all x ΠR, then
(a) m < 5
(b) m > 5
(c) m < 71/24
(d) m > 71/24
60. If a, b, c are distinct real numbers, then the quadratic expression
( x - b) ( x - c) ( x - c) ( x - a ) ( x - a ) ( x - b)
+
+
( a - b ) ( a - c ) (b - c ) (b - a ) (c - a ) (c - b )
is identically equal to
Quadratic Equations 3.35
(a) 1
(c) x2
69. The number of real solution of 2 sin
= 7 is
(b) x
(d) none of these
61. If ax2 + bx + c, a, b, c Œ R, a π 0 has no real zero
and a – b + c < 0, then value of ac is
(a) positive
(c) negative
(b) zero
(d) non-negative
62. Suppose p ΠR. Let a, b be roots of x2 Р2x Р(p2
– 1)= 0 and g, d (g < d ) be roots of x2 – 2(p + 1)
x + p(p Р1) = 0. If a, b Π(g, d ), then p lies in
the interval
(a) (–1/4, 1)
(b) (–1, 1)
(c) (0, •)
(d)
(
2, • )
63. If a, b are the roots of the equation ax2 + 2bx + c
= 0 and a + h, b + h are the roots of the equation
Ax2 + 2Bx + C = 0, then
b2 - ac
a
(a) 2
=
B - AC A
(c) h =
bA - aB
2 Aa
2
2
64. The quadratic equation x + 7x = 14 (q + 1), where
q is an integer has
(a)
(b)
(c)
(d)
real and distinct roots
integral roots
imaginary roots
none of these
65. Let a, b, c ΠR and a > 0. If the quadratic equation
ax2 + bx + c = 0 has two real roots a and b such
c b
that a < – 1 and b > 1, then value of +
is
a a
(a) less than 2
(b) less than 1
(c) less than 0
(d) less than – 1
66. Let a, b, c Œ R and a π 0 be such that (a + c)2 <
b2, then the quadratic equation ax2 + bx + c = 0 has
(a)
(b)
(c)
(d)
imaginary roots
real roots
two real roots lying between (– 1, 1)
none of these
(x – a) (x – 10) + 1 = 0 has integeral roots are
(a) – 1, 3
(b) 2, 3
(c) 12, 8
(d) – 8, – 12
68. The number of real solution of 4
(10) (6x) is
(a) zero
(c) two
(b) one
(d) infinite
(a) 0
(c) 2
(b) 1
(d) infinitely many
71. If roots of the quadratic equation ax2 + bx + c = 0
a a +1
are real and are of the form
, then value
,
a -1 a
2
of (a + b + c) is
(a) 4ac – b2
(c) c2 + a2 – 2b2
+ 9
x + 0.5
(b) b2 – 4ac
(d) none of these
(b) 1
(d) any of (a), (b) and (c)
73. If a, b are the roots of the equation x2 + ax + b = 0,
1
then maximum value of – x2 + ax + b + (a –
4
b)2 is
(a)
1 2
(a – 4b)
4
(b)
(c)
a2
2
(d) none of these
(a) b = – 1, c = 2
(c) b = – 5, c < 2
(b) b > – 2, c < 1
(d) none of these
75. Let a, b, c ΠR be such that a + b + c < 0, a Рb +
c < 0 and c > 0. If a and b are roots of the equation
ax2 + bx + c = 0, then value of [a] + [b] is
(b) 1
(d) 0
76. If roots of the equation x2 – 2mx + m2 – 1 = 0 lie
in the interval (– 2, 4), then
(a) – 1 < m < 3
(c) 1 < m < 3
77. The value of
=
1 2
(b – 4a)
4
74. If both the roots of the equation x2 + bx + c = 0
lie in the interval (0, 1), then
(a) 2
(c) – 1
67. The integral values of a for which the quadratic
equation
x + 1.5
(b) 1
(d) infinitely many
70. The number of values of k for which the equation
x2 – 2x + k = 0 has two distinct roots lying in the
interval (0, 1) is
(a) 0
(c) 2
Ac + aC
Aa + Bb
2
2
+ 5(2 cos x )
72. Let x be an integer and x2 + x + 1 is divisible by
3. When x is divided by 3, it leaves remainder
2
(b) b – ac = B – AC
(d) h =
(a) zero
(c) finitely many
2x
(a) 10
(c) 8
(b) 1 < m < 5
(d) – 1 < m < 5
8 + 2 8 + 2 8 + 2 8 + is
(b) 6
(d) none of these
78. The number of solutions of the equation
Ê px ˆ
= x2 – 2 3 x + 4 is
sin Á
Ë 2 3 ˜¯
3.36
Complete Mathematics—JEE Main
(a) 1
(c) 0
(b) 2
(d) infinite
79. The number of solutions of |x + 2| = 2(3 – x) is
(a) 1
(c) 3
(b) 2
(d) 0
80. If the equation ax2 + bx + c = 0, a > 0 has two
distinct roots a and b such that a < – 2 and b >
3, then
(a) c > 0
(c) c < 0
(b) c = 0
(d) c = a – b
81. Two non-integer roots of
(x2 – 5x)2 – 7(x2 – 5x) + 6 = 0
are
(a) 0
(c) 3
(b) 2
(d) 4
87. The number of irrational roots of the equation
1ˆ
1 2
Ê
4 ÊÁ x - ˆ˜ + 8 Á x + ˜ = 29
Ë
Ë
¯
x¯
x
is
(a) 0
(b) 2
(c) 4
(d) infinite
88. Irrational roots of the equation
2x4 + 9x3 + 8x2 + 9x + 2 = 0
are
(a) - 2 - 3 , 2 + 3
(b) 2 - 3 , 2 + 3
(c) - 2 + 3 , - 2 - 3 (d) none of these
(a)
(b)
(c)
(d)
1
1
(5 + 29 ), (5 - 29 )
2
2
1
1
(- 5 + 29 ), (- 5 + 29 )
2
2
1
1
(- 5 + 14 ), (- 5 - 41)
2
2
none of these
89. Sum of the roots of the equation
is
(a) 5
(c) – 5/2
2
Ê x - 1ˆ
Ê x - 1ˆ
ÁË x + 1˜¯ – 13 ÁË x + 1˜¯ + 36 = 0, x π – 1
83. The number of negative roots of
9x + 2 – 6(3x + 1) + 1 = 0
(b) 1
(d) 4
(a) 1
(c) 3
is
(a) 2
(c) 7
2
+ 4 = 0, x π 1/3
(x2 + 3x + 2)2 – 8(x2 + 3x) – 4 = 0
(b) 2
(d) 4
86. The number of roots of the equation
is
3x - 6 = 2
(b) 5
(d) 10
92. The number of real roots of
85. The number irrational roots of
x
+
x-3
(b) 2
(d) 4
x–
2 2 x + 1 = 2x – 1
is
(a) 1
(c) 3
x-3
5
= , x π 0, x π 3
2
x
(b) 2
(d) 4
93. Product of roots of the equation
(b) 2
(d) 4
is
(a) 0
(c) 3
(b) 1
(d) – 1
is
(a) 0
(c) 3
84. The number of rational roots of
4
1ˆ
˜ +1=0
x¯
91. Product of roots of the equation
(b) 2
(d) 4
Ê 2 x - 5ˆ
Ê 2 x - 5ˆ
81 Á
– 45 Á
Ë 3 x + 1 ˜¯
Ë 3 x + 1 ˜¯
is
Ê
ÁË x -
(x – 1) (x – 2) (3x – 2) (3x + 1) = 21
4
is
(a) 0
(c) 2
1ˆ 2
˜ –4
x¯
90. The number of irrational roots of the equation
82. The number of real roots of
is
(a) 0
(c) 3
Ê
4Áx Ë
13 - x 2 = x + 5
is
(a) – 6
(c) 6
(b) 7
(d) – 7
94. The number of roots of the equation
x 2 - 4 - ( x - 2) = x 2 - 5 x + 6 is
(a) 0
(b) 1
(c) 2
(d) 3
95. The product of the roots of the equation
Quadratic Equations 3.37
98. If a is a root of x4 + x2 – 1 = 0, the value of
(a6 + 2a4)2012 is
x 2 - 4 x + 3 + x 2 - 7 x + 12 = 3 x - 3 is
(a) 15
(b) – 15
(c) 20
(d) – 20
96. Suppose a and b satisfy the equations
18a2 + 77a + 2 = 0 and 2b2 + 77b + 18 = 0 then
ab + a + 1
is
value of
b
(a) – 25
(b) – 25/6
(c) 6/25
(d) – 1/25
97. Suppose a, b are roots of x2 – 7x + 8 = 0, with a
16
+ 3b 2 - 19b is
> b, then value of
a
(a) – 10
(b) 10
(c) – 23
(d) 17
(a) 0
(b) – 1
(c) 1
(d) none of these
99. Sum and product of all the roots of the equation
(x2 – x – 1) (x2 – x – 2) (x2 – x – 3) … (x2 – x –
2012) = 0 is
(a) 2012, 2012!
(b) – 2012, 2012!
(c) – 2012, – 2012!
(d) 2012, – 2012!
100. Suppose three distinct non-zero real numbers satisfy
a2(a + k) = b2(b + k) = c2(c + k), where k is some
1 1 1
real number, then value of + + is
a b c
(a) 0
(b) k
(c) – k
(d) 2k
Assertion-Reason Type Questions
101. Suppose a, b, c, a, b Œ R a π 0 and 0 < a < b.
Let f(x) = a2x2 + bx + c, g(x) = a2x2 – bx – c
Statement-1: If ax2 + bx + c = x does not have
real roots, then
Statement-1: If a is a root of f(x) = 0 and b is a
root of g(x) = 0, then there exists g Π(a, b ) such
that g is a root of a2x2 + 2bx + 2c = 0.
a(ax2 + bx + c)2 + b(ax2 + bx + c) + c = x has two
real and two imaginary roots.
Statement-2: If a function h : [a, b] Æ R is continuous on [a, b] and h(a) h(b) < 0, then there
exists g Π(a, b) such that h(g ) = 0.
102. Suppose p ΠR and f(x) = x2 + 2px + p
Statement-1: If f(x) < 0 " x Π[1, 2] then
p Œ (– •, –4/5).
Statement-2: a lies between the roots of f(x) = 0,
if and only if f(a) < 0.
106. Suppose a, b, c Œ R, a π 0 and a – b + c < 0.
Statement-1: If ax2 + bx + c = 0 has imaginary
roots, and
yˆ
Ê
a Ë x 2 + ¯ + (b + 1) x + c = ax 2 + bx + c + x + y
a
then (x, y) lies in the half plane {(x, y)| x + y £ 0}
Statement-2: |a + b| = |a| + |b| ¤ ab ≥ 0
103. Suppose a ΠR.
Statement-1: The equation a (2 x - 2 ) + 1 = 1 – 2x
has a real solution for all a Π(0, 1].
Statement-2: ax2 + bx + c = 0, a π 0 has two
positive roots if ab > 0 and ac > 0.
(
104 Suppose a, b, c ŒR and f(x) = 2(a – x) x + x 2 + b2
x ΠR.
2
)
2
Statement-1: Maximum value of f(x) is a + b .
Statement-2: If a < 0, the maximum value of f(x)
= ax2 + bx + c is (b2 – 4ac)/4a.
105. Suppose a, b, c Œ R, a π 0.
Statement-2: If a + ib, where a, b Œ R, b π 0 is
a root of ax2 + bx + c = x, then its other root must
be a – ib
107. a, b, c ΠR and c < min {a, b}.
( x + a ) ( x + b)
, x > – c.
Let f(x) =
x+c
Statement-1: Minimum value of f is
(
2
a - c + b - c)
Statement-2: If a < 0, then ax2 + bx + c has no
minimum value.
108. Suppose a, b, c, p, q, r Œ R, a, p π 0.
Let f(x) = ax2 + bx + c and g(x) = px2 + qx + r
Statement-1: If f(x) = g(x) for three distinct real
values of x then a = p, b = q and c = r.
3.38
Complete Mathematics—JEE Main
Statement-2: If a π p, then f(x) – g(x) = 0 has two
real roots.
110. Suppose a, b, c Œ R a π 0 and c > 0.
Let f(x) = ax2 + bx + c.
Statement-1: If f(x) = 0, does not have real roots,
then
¸
Ï 2
a > max Ì b , b - c ˝
˛
Ó 4c
2
Statement-2: If b – 4ac < 0 then f(x) > 0 " x
ΠR.
109. Suppose p ΠR.
Statement-1: If sin4 x + psin2 + 1 = 0 has a solution, then p Œ (– •, –2]
Statement-2: y +
1
≥ 2 " y > 0.
y
LEVEL 2
Straight Objective Type Questions
111. If a Π(0, p/2), then the expression
1
1
1
+
+
= 0, has
x – sin a x – sin b x – sin g
2
x2 + x +
tan a
x2 + x
is always greater than or equal to
(a) 2 tan a
(b) 2
(c) 1
(d) sec2 a
112. If a, b ΠR, and the equation
x2 + (a – b)x – a – b + 1 = 0
has real roots for all b ΠR, then a lies in the interval
(a) (1, •)
(b) (0, •)
(c) (– •, 1)
(d) (– 1, 1)
118.
119.
113. If a and b (a < b) are the roots of the equation
x2 + bx + c = 0, where c < 0 < b, then
(a) 0 < a < b
(c) a < b < 0
(b) a < 0 < b < |a |
(d) none of these.
120.
114. Suppose a, b, c ΠR and the equation x2 +
(a + b)x + c = 0 has no real roots, then which one
of the following is not true.
(a)
(b)
(c)
(d)
c + c (a + b + c) > 0
c – c (a + b – c) > 0
(a + b)2 – 4c < 0
c (a + b + c) < 0
121.
115. If [x] denotes the greatest integer £ x, then number
of solutions of the equation x2 – 2 – 2[x] = 0 is
(a) 4
(c) 3
(b) 2
(d) none of these
116. If [x] denotes the greatest integer £ x, and a, b
are two odd integers, then number of solutions of
[x]2 + a[x] + b = 0 is
(a) 1
(c) 2
(b) 0
(d) infinite
117. Let a, b, g be distinct real numbers lying in
(0, p/2), then the equation
122.
(a) two distinct real roots
(b) two equal roots
(c) two imaginary roots
(d) one real and one imaginary root.
If roots of the equation x2 + ax + b = 0 are
a, b, then the roots of x2 + (2a + a)x + a2 +
aa + b = 0 are
(a) 1, b – a
(b) 0, a – b
(c) 0, b – a
(d) 0, 1.
Let a, b, c be the sides of a triangle with a π c and
l ΠR. If the roots of x2 + 2(a + b + c)x + 3l (ab
+ bc + ca) = 0 are real, then l lies in
(a) (– •, 4/3)
(b) (5/3, •)
(c) (1/3, 5/3)
(c) (4/3, 5,3)
If tan q and cot q are roots of x2 + 2ax + b = 0,
then least value of |a| is
1
(b) 1
(a)
2
(c) 2
(d) cannot be found.
Let a, b, g be the roots of x3 + x2 – 5x – 1 = 0,
then value of [a] + [b] + [g], where [x] denotes the
greatest integer £ x, is
(a) 1
(b) 2
(c) – 2
(d) – 3.
[Hint: Show that f (–3) < 0, f (– 2) > 0, f (– 1) >
0, f (0) < 0, f (1) < 0, f (2) > 0]
If tan A and tan B are roots of the quadratic equation
x2 – px + q = 0, then the value of sin2 (A + B) is
(a)
(c)
p2
p2 + q 2
p2
(1 – q )2 + p2
(b)
p2
(q + p )2
(d) 1 –
p2
(1 – q )2
Quadratic Equations 3.39
123. The equation
x + 3 + 4 x – 1 + x + 8 + 6 x – 1 = 1 has
(a) no solution
(b) only one solution
(c) only two solutions
(d) infinite number of solutions
124. The number of irrational solutions of the equation
x 2 + x 2 + 11 + x 2 – x 2 + 11 = 4 is
(a) 0
(b) 2
(c) 4
(d) infinite
125. Let a, b, c, p, q be five different non-zero real
numbers and x, y, z be three numbers satisfying the
system of equations
x
y
z
+
+
=1
a a– p a–q
and
x
y
z
+
+
=1
b b– p b–q
x
y
z
+
+
=1
c c– p c–q
then x equals
abc
pq
(a)
(b)
pq
abc
abc
(c)
(d) none of these.
p+q
126. Let f (x) = ax2 + bx + c, where a, b, c ΠR.
Suppose |f (x)| £ 1 " x Œ [0, 1], then |a| cannot
exceed
(a) 5
(b) 6
(c) 7
(d) 8
127. If a(p + q)2 + 2bpq + c = 0 and
a(p + r)2 + 2bpr + c = 0,
then |q – r| equals
(a)
2
a
(2a + b)bp2 - ac
(b)
2
a
p2 - 4ac
(c) p2 + c
(d) none of these.
a
128. Let a, b, c Œ R be such that b2 ≥ 4ac. All the four
roots of the equation ax4 + bx2 + c = 0 will be real
if
(a) a > 0, b < 0, c > 0 or a < 0, b > 0, c < 0
(b) a > 0, b > 0, c > 0 or a < 0, b < 0, c < 0
(c) a > 0, b < 0, c > 0 or a > 0, b > 0, c < 0
(d) none of these.
129. If the equations
x2 + mx + 1 = 0 and
(b – c) x2 + (c – a) x + (a – b) = 0 have a common
root, then
(a) m = – 2
(b) m = – 1
(c) m = 0
(d) m = 1
130. If x, y, z ΠR, x + y + z = 4 and x2 + y2 + z2 = 6,
then the maximum possible value of z is
(a) 1
(b) 2
(c)
3
2
(d)
4
3
Previous Years' AIEEE/JEE Main Questions
1. If a π b and a2 = 5a – 3, b2 = 5b – 3, then the
a
b
equation whose roots are
and
is
b
a
(a) 3x2 – 25x + 3 = 0 (b) x2 – 5x + 3 = 0
(d) 3x2 – 19x + 3 = 0
(c) x2 + 5x – 3 = 0
[2002]
2. If a π b and differences between the roots of the
equations x2 + ax + b = 0 and x2 + bx + a = 0 is
the same, then
(a) a + b + 4 = 0
(b) a + b – 4 = 0
(c) a – b + 4 = 0
(d) a – b – 4 = 0
[2002]
3. If a, b, c ΠR and 2a + 3b + 6c = 0, then the
equation ax2 + bx + c = 0 has
(a) at least one root in [0, 1]
(b) at least one root in [2, 3]
(c) at least one root in [– 1, 0]
(d) at least one root in (– •, 1)
[2002]
4. Product of real roots of the equation t2x2 + |x| +
9 = 0, t ΠR is always
(a) positive
(b) negative
(c) zero
(d) does not exit
[2002]
5. The value of a for which one root of the quadratic
equation
(a2 – 5a + 3)x2 + (3a – 1)x + 2 = 0
is twice as large as the other is
2
1
(b)
(a) –
3
3
1
2
(c) –
(d)
[2003]
3
3
Complete Mathematics—JEE Main
3.40
6. If sum of the roots of the quadratic equation ax2
7.
8.
9.
10.
11.
12.
+ bx + c = 0 is equal to the sum of the squares of
a b
c
their reciprocals, then ,
and
are in
c a
b
(a) G.P.
(b) H.P.
(c) A.G.P.
(d) A.P.
[2003]
The number of real solution of the equation x2 –
3|x| + 2 = 0 is
(a) 4
(b) 1
(c) 3
(d) 2
[2004]
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are roots of the
quadratic equation
(a) x2 + 18x – 16 = 0 (b) x2 – 18x + 16 = 0
(c) x2 + 18x + 16 = 0 (d) x2 – 18x – 16 = 0
[2004]
If 1 – p is a root of quadratic equation x2 + px +
1 – p = 0, then its roots are
(a) 0, – 1
(b) – 1, 1
(c) 0, 1
(d) – 1, 2
[2004]
If one root of the equation x2 + px + 12 = 0 is 4,
while equation x2 + px + q = 0 has equal roots,
then the value of q is
(a) 3
(b) 12
49
(d) 4
[2004]
(c)
4
If 2a + 3b + 6c = 0, then at least one root of the
equation ax2 + bx + c = 0 lies in the interval
(a) (2, 3)
(b) (1, 2)
(c) (0, 1)
(d) (1, 3)
[2004]
p
P
. If tan Ê ˆ and
In a triangle PQR, –R =
Ë2 ¯
2
Q
tan Ê ˆ are the roots of ax2 + bx + c = 0, a π 0,
Ë2 ¯
then
(a) b = c
(b) b = a + c
(c) a = b + c
(d) c = a + b
[2005]
13. The value of a for which sum of the squares of the
roots of the equation of the roots of the equation
x2 – (a – 2)x – a – 1 = 0
assume least value is
(a) 3
(b) 2
(c) 1
(d) 0
[2005]
2
14. If the roots of the equation x – bx + c = 0 be two
consecutive integers then b2 – 4c equals
(a) 2
(b) 1
(c) – 2
(d) 3
[2005]
15. If both the roots of the quadratic equation x2 – 2kx
+ k2 + k – 5 = 0 are less than 5, then k lies in the
interval
(a) (– •, 4)
(b) [4, 5]
(c) (5, 6)
(d) (6, •)
[2005]
16. If the equation anxn + an – 1xn – 1 + º + a1x = 0,
a1 π 0, n ≥ 2 has a positive root x = a, then the
equation
n an xn – 1 + (n – 1) an – 1 xn – 2 + º + a1 = 0
has a positive root, which is
(a) greater than or equal to a
(b) equal to a
(c) greater than a
(d) smaller than a
[2005]
2
17. If the roots of the quadratic equation x + px + q
= 0 are tan 30° and tan 15° respectively, then the
value of 2 + q – p is
(a) 1
(b) 2
(c) 3
(d) 0
[2006]
18. All the values of m for which both the roots of the
equation x2 – 2mx + m2 – 1 = 0 are greater than – 2
but less than 4, lie in the interval
(a) 1 < m < 4
(b) – 2 < m < 0
(c) m > 3
(d) – 1 < m < 3
[2006]
3 x 2 + 9 x + 17
19. If x is real, the maximum value of
3x2 + 9 x + 7
is
17
1
(b)
(a)
7
4
(c) 41
(d) 1
[2006]
20. If the difference between the roots of x2 + ax + 1
= 0 is less than 5 , then set of possible values of
a lie in the interval
(a) (– 3, 3)
(b) (– 3, •)
(c) (3, •)
(d) (– •, – 3)
[2007]
21. The quadratic equations x2 – 6x + a = 0 and x2 – cx
+ 6 = 0 have one common root. The other roots of
first and second equation are integers in the ratio 4
: 3. Then common root is
(a) 1
(b) 4
(c) 3
(d) 2
[2008]
22. If the roots of the equation bx2 + cx + a = 0 be
imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is
(a) greater then – 4ab (b) less then – 4ab
(c) greater than 4ab (d) less than 4ab
[2009]
23. If a and b are the roots of the equation x2 – x + 1
= 0, then a2009 + b2009 equals
(a) 1
(b) 2
(c) – 2
(d) –1
[2010]
24. Let for a, a1 π 0, a π a1
f(x) = ax2 + bx + c, g(x) = a1x2 + b1x + c1 and
p(x) = f(x) – g(x).
If p(x) = 0 only for x = –1 and p(–2) = 2, then
value of p(2) is:
(a) 3
(b) 9
(c) 6
(d) 18
[2011]
Quadratic Equations 3.41
25. The equation esin x – e– sin x = 4 has
(a) no real roots
(b) exactly one real root
(c) exactly four real roots
(d) infinite number of real roots.
[2012]
2
2
26. If the equations x + 2x + 3 = 0 and ax + bx + c = 0,
a, b, c ΠR have a common root, then a: b: c is
(a) 3 : 2 : 1
(b) 1 : 3 : 2
(c) 3 : 1 : 2
(d) 1 : 2 : 3
[2013]
27. If a and b are the roots of the equation x2 + px +
3
p = 0, such that |a – b | = 10 , then p belongs
4
to the set
(a) {2, – 5}
(b) {–3, 2}
(c) {–2, 5}
(d) {– 3, 5} [2013, online]
28. If p and q are non-zero real number such that a3
+ b 3 = – p and ab = q, then a quadratic equation
a2
b2
whose roots are
and
is
b
a
(a) px2 – qx + p2 = 0
(b) qx2 + px + q2 = 0
(c) px2 + qx + p2 = 0
[2013, online]
(d) qx2 – px + q2 = 0
29. The values of a for which one root of the equation
x2 – (a + 1)x + a2 + a – 8 = 0 exceeds 2 and the
other is less than 2 are given by
(a) 3 < a < 10
(b) a > 10
(c) –2< a < 3
(d) a £ –2 [2013, online]
30. The least integral value a of x such that
x-5
> 0, satisfies
2
x + 5 x - 14
(a) a 2 + 3a – 4 = 0
(c) a 2 – 7a + 6 = 0
(b) a 2 – 5a + 4 = 0
(d) a 2 + 5a – 6 = 0
[2013, online]
2
31. Let a and b be the roots of the equation px + qx
1 1
+ r = 0, p π 0. If p, q, r are in A.P. and
+
=
a b
4, then value of |a – b | is
1
2
(a)
(b)
61
17
9
9
1
2
34
13
(d)
[2014]
9
9
32. If equations ax2 + bx + c = 0, (a, b,c Œ R, a π 0)
and 2x2 + 3x + 4 = 0 have a common root, then
a : b : c equals:
(c)
(a) 1 : 2 : 3
(c) 4 : 3 : 2
(b) 2 : 3 : 4
(d) 3 : 2 : 1 [2014, online]
2
33. The sum of the roots of the equation x + |2x – 3|
+ 4 = 0, is:
(a) 2
(c)
(b) –2
(d) - 2
2
[2014, online]
2
34. If a and b are the roots of x – 4 2 kx + 2e4lnk
– 1 = 0 for some k, and a2 + b 2 = 66, then a 3 +
b 3 is equal to:
(a) 248 2
(b) 280 2
(c) -32 2
(d) -280 2 [2014, online]
35. The equation 3 x 2 + x + 5 = x – 3, where x is
real, has:
(a) has no solution
(b) exactly one solution
(c) exactly two solutions
(d) exactly four solutions
[2014, online]
36. Let a and b be the roots of equation x2 – 6x – 2
= 0. If an = a n – b n, for n ≥ 1, then the value of
a10 - 2a8 is equal to:
2a9
(a) 6
(b) – 6
(c) 3
(d) –3
[2015]
37. If 2 + 3i is one of the roots of the equation 2x3 –
9x2 + kx Р13 = 0, k ΠR, then the real root of the
equation:
(a) does not exist.
(b) exists and is equal to 1
2
(c) exists and is equal to – 1
2
(d) exists and is equal to 1
[2015, online]
38. If the two roots of the equation, (a – 1)(x4 + x2 +1)
+ (a + 1)(x2 + x + 1)2 = 0 are real and distinct, then
the set of all values of a is:
(a) ÊÁ - 1 , 0ˆ˜
Ë 2 ¯
(b) (–•, –2) » (2, •)
1
1
1
(c) ÊÁ - , 0ˆ˜ » ÊÁ 0, ˆ˜ (d) ÊÁ 0, ˆ˜
Ë 2¯
Ë 2 ¯ Ë 2¯
[2015, online]
39. The sum of all real values of x satisfying the equa2
tion ( x 2 - 5 x + 5) x + 4 x - 60 = 1 is
(a) 3
(b) –4
(c) 6
(d) 5
[2016]
40. If the equations x2 + bx – 1 = 0 and x2 + x + b =
0 have a common root different from –1, then |b|
is equal to:
(a) 2
(b) 3
(d) 2
[2016, online]
(c) 3
41. If x is a solution of the equation, 2 x + 1 - 2 x - 1
1ˆ
Ê
= 1, Á x ≥ ˜ , then 4 x 2 - 1 is equal to:
Ë
2¯
3.42
Complete Mathematics—JEE Main
(a)
3
4
(b)
(c) 2 2
1
2
(d) 2
[2016, online]
42. Let x, y, z be positive real numbers such that x +
y + z = 12 and x3y4z5 = (0.1)(600)3. Then x3 + y3
+ z3 is equal to
(a) 342
(b) 216
(c) 258
(d) 270
[2016, online]
Previous Years' B-Architecture Entrance
Examination Questions
8. If the quadratic equation 3x2 + 2(a2 + 1)x + a2 –
3a + 2 = 0 posseses roots of opposite signs, then
a lies in the interval:
1. If the roots of the quadratic equation x2 + 2px + q
= 0 are tan 30° and tan 15° respectively, then q is
(a) 1 + p
(c) 1 – 2p
(b) 1 – p
(d) 1 + 2p
[2006]
2. The set of values of a for which the quadratic
equation (a + 2) x2 – 2ax – a = 0 has two roots
on the number line symmetrically placed about the
point 1 is
(a) {–1, 0}
(c) f
(b) {0, 2}
(d) {0, 1}
[2007]
2
3. The number of solutions of the equation x – 4|x|
– 2 = 0 is:
(a) 1
(c) 3
(b) 2
(d) 4
[2008]
4. The quadratic equation whose roots are a/b and b/a,
a π b π 0, where a2 = 5a – 3, and b2 = 5b – 3, is:
(a)
(b)
(c)
(d)
3x2 – 19x + 3 = 0
3x2 + 19x – 3 = 0
3x2 + 19x + 3 = 0
3x2 – 19x – 3 = 0
[2009]
2
5. If the roots of the quadratic equation ax + bx + c =
0 are a, b, then the roots of the quadratic equation
ax2 – bx(x – 1) + c(x – 1)2 = 0, are
a +1 b +1
a
b
,
(b)
,
(a)
a
b
a -1 b -1
(c)
a
b
,
a +1 b +1
(d) 1 – a, 1 – b
[2010]
6. If x2 – 3x + 2 is a factor of x4 – ax2 + b = 0 then
the equation whose roots are a and b is
(a) x2 + 9x + 20 = 0
(c) x2 – 9x + 20 = 0
(b) x2 – 9x – 20 = 0
(d) x2 + 9x – 20 = 0 [2011]
7. Let a, b, c Œ R, a > 0 and the function f : R Æ
R be defined by f(x) = ax2 + bx + c.
Statement-1: b2 < 4ac fi f(x) > 0 for every value
of x.
Statement-2: f is strictly decreasing in the interval
(– •, – b/2a) and strictly increasing in the interval
(–b/2a, •).
[2012]
(a) (– •, –1)
(c) (1, 2)
(b) (–1, 1)
(d) (2, 3)
[2013]
1
1
1
9. If the roots of the equation
+
= are
x+ p x+q r
equal in magnitude and opposite in sign, then product of roots is:
1 2
(b)
(p + q2)
(a) p2 + q2
2
1 2
1 2
(c) –
(p + q2)
(d) –
(p – q2)
[2014]
2
2
10. The values of k for which each root of the equation, x2 – 6kx + 2 – 2k + 9k2 = 0 is greater than 3,
always satisfy the inequality:
(a) 7 – 9y > 0
(b) 11 – 9y < 0
(c) 29 – 11y > 0
(d) 29 – 11y < 0
[2015]
11. The number of integral values of m for which the
equation, (1 + m2)x2 – 2(1 + 3m)x + (1 + 8m) = 0,
has no real root, is:
(a) 2
(b) 3
[2016]
Answers
Concept-based
1.
5.
9.
13.
17.
(a)
(b)
(d)
(b)
(c)
2.
6.
10.
14.
18.
(c)
(b)
(b)
(b)
(a)
3.
7.
11.
15.
19.
(c)
(a)
(c)
(d)
(a)
4.
8.
12.
16.
20.
(d)
(a)
(c)
(a)
(a)
22.
26.
30.
34.
38.
(c)
(b)
(a)
(a)
(b)
23.
27.
31.
35.
39.
(c)
(a)
(d)
(b)
(d)
24.
28.
32.
36.
40.
(b)
(c)
(d)
(b)
(a)
Level 1
21.
25.
29.
33.
37.
(b)
(a)
(a)
(a)
(a)
Quadratic Equations 3.43
41. (d)
45. (b)
49. (c)
53. (a)
57. (a)
61. (a)
65. (d)
69. (d)
73. (c)
77. (d)
81. (a)
85. (d)
89. (b)
93. (c)
97. (a)
101. (a)
105. (d)
109. (a)
42.
46.
50.
54.
58.
62.
66.
70.
74.
78.
82.
86.
90.
94.
98.
102.
106.
110.
(a)
(c)
(b)
(d)
(c)
(a)
(b)
(a)
(d)
(a)
(d)
(b)
(b)
(d)
(c)
(a)
(a)
(a)
43.
47.
51.
55.
59.
63.
67.
71.
75.
79.
83.
87.
91.
95.
99.
103.
107.
112.
116.
120.
124.
128.
(a)
(b)
(b)
(b)
(a)
113.
117.
121.
125.
129.
(b)
(d)
(b)
(b)
(c)
(c)
(c)
(b)
(c)
(a)
(b)
(b)
(d)
(a)
(a)
(c)
(b)
44.
48.
52.
56.
60.
64.
68.
72.
76.
80.
84.
88.
92.
96.
100.
104.
108.
(d)
(a)
(b)
(a)
(a)
(a)
(c)
(b)
(a)
(c)
(d)
(c)
(a)
(b)
(a)
(c)
(b)
Hints and Solutions
Concept-based
1. (x – 1)2 + 1 + |x – 1| ≥ 1 " x Œ R
k
Ê 1ˆ
2. 1 = a Ë ¯ =
a
7
3. a – b = –1 = fi (a + b )2 – 4a b = 1 or p2 – 4q = 1
4. If a is a common root, then a 2 – aa + b = 0, a2 +
ba – a = 0
Subtracting, we get – (a + b) a + b + a = 0
fi a = 1. Thus, 1 – a + b = 0 fi a – b = 1
Roots of the equation are –
These will exceed a, if -
Level 2
111. (a)
115. (d)
119. (a)
123. (a)
127. (a)
(b)
(a)
(d)
(a)
(a)
114.
118.
122.
126.
130.
(d)
(c)
(c)
(d)
(b)
Previous Years' AIEEE/JEE Main Questions
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
(d)
(d)
(a)
(c)
(c)
(d)
(a)
(c)
(c)
(b)
(a)
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
(a)
(b)
(c)
(b)
(d)
(a)
(d)
(d)
(b)
(c)
(b)
2
1ˆ
1
= –a
¯
2
4
1
1
We must have
–a≥0fia£ .
4
4
Ê
5. x2 + x + a = 0 fi Ëx +
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
(a)
(a)
(c)
(a)
(c)
(a)
(c)
(d)
(a)
(a)
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
(d)
(b)
(d)
(d)
(a)
(d)
(b)
(b)
(c)
(c)
Previous Years' B-Architecture Entrance
Examination Questions
fi -
1
1
±
-a
2
4
1
1
-a >a
2
4
1
1
-a >a+
4
2
This is possible if a + 1/2 < 0 and
1
1ˆ
Ê
- a < Ëa + ¯
4
2
1
1
– a < a2 + a +
4
4
2
fi a + 2a > 0 fi a < – 2 as a < 0.
fi
6. (|x| – 2)2 = 6 fi |x| = 2 ± 6 .
As |x| ≥ 0, |x| = 2 +
6 fi x = ± (2 +
6)
2
7. 10 = ( 10 ) = |a – b |2 = (a + b)2 – 4ab = p2 – 3p
fi p2 – 3p – 10 = 0 fi (p – 5) (p + 2 ) = 0
fi p = – 2, 5
8. For the equation to be defined, x ≥ 0, 5 + x ≥ 0. For
x ≥ 0, x + 5 ≥ 5 fi x + x + 5 ≥ 5 > 2
9. D = 4(a + 1)2 – 4(a2 – 4a + 3) ≥ 0,
2(a + 1) < 0 and a2 – 4a + 3 > 0
First two imply a ≥ 1/3, a < – 1. Not possible.
10. tan a + tan b = – a, tan a tan b = b.
tan a + tan b
a
=
1 - tan a tan b 1 - b
Also, 0 < a < p /4, 0 < b < p /4
1 = tan (a + b) =
1. (d)
2. (c)
3. (b)
4. (a)
5. (c)
6. (c)
7. (a)
8. (c)
9. (c)
10. (b)
11. (c)
3.44
11.
12.
13.
14.
Complete Mathematics—JEE Main
fi 0 < tan a < 1, 0 < tan b < 1
fi 0 < tan a tan b < 1 fi 0 < b < 1.
Let f(x) = (x – a) (x – b) – c
Use coefficient of x2 > 0,
f(a) = – c = f(b) < 0.
D ≥ 0 and product of roots £ 0.
fi (k + 1)2 – (9k – 5) ≥ 0 and 9k–5 £ 0
fi k2 – 7k + 6 ≥ 0 and £ 5/9
fi (k – 1) (k – 6) ≥ 0 and k £ 5/9
Thus, k £ 5/9
Use : a and a (k – a) (k – b) have the opposite signs,
where a, b are roots of
ax2 + bx + c = 0.
Write the equation as
(x – 2a)2 = 4a2 – (2a2 – 3a + 5)
= (2a + 5) (a – 1)
fi x = 2a ± (2a + 5) ( a - 1)
We must have (2a + 5) (a – 1) ≥ 0
and 2a + (2a + 5) ( a - 1) < 2
fi a £ – 5/2 or a ≥ 1
and
15.
16.
17.
18.
19.
(2a + 5) (a - 1) < – 2(a – 1)
Clearly, a ≥ 1 is not possible
For a £ – 5/2, we must have
- (2a + 5) (1 - a ) < 2 (1 – a)
fi – (2a + 5) < 4(1 – a) fi a < 9/2
As parabola open upwards, a > 0.
Also, for x = 0 ax2 + bx + c = c < 0.
Since the equation has distinct roots, b2 – 4ac > 0.
However, we cannot be sure of sign of b.
If can be positive, negative or zero. For instance, consider x2 + 2x – 3 = 0, x2 – 2x – 3 = 0 and x2 – 3 = 0.
9x2 – 6x + 5 = (3x – 1)2 + 4 attains least value when
x = 1/3.
ax2 + bx + c = 0 has real roots if
b2 – 4ac ≥ 0, i.e. if b2 ≥ 4ac.
Minimum possible value of b is 2.
For b = 2, (a, c) = (1, 1)
For b = 3, ac £ 9/4 fi (a, c) = (1, 1), (1, 2) or (2, 1)
For b = 4, ac £ 4 fi (a, c) = (1, 1), (1, 2), (2, 1),
(1, 3), (3, 1), (1, 4), (4, 1), (2, 2)
Thus, there 43 – 12 = 52 such quadratic equations.
LHS > 0 " x ΠR.
As the roots are imaginary b2 – 4ac < 0 and
1
- b ± i 4ac - b2
a, b =
2a
(
|a | =
1
b2 + 4 ac - b2 =
2a
)
c
= |b|
a
20. Suppose a, b > 1, a < b be roots of ax2 + bx + c = 0, then
ax2 + bx + c > 0 if x < a or x > b
< 0 if a < x < b
Thus, a + b + c > 0.
Level 1
21. a > 0, c > 0, 4b2 – 4ac < 0. However nothing can be
said about b. For instance, consider parabolas y = x2
– x + 1, y = x2 + x + 1 or y = x2 + 1
22. If a is a common root, then aa2 + 2ba + c = 0, aa2 +
2ca + b = 0
Subtracting, we get 2(b – c) a + c – b = 0
1
1
fi a = . Thus a + b + c = 0
2
4
a+b+c 3
fi
=
a
4
2
23. Let f(x) = ax + bx + c. As f(x) = 0, does not have real
roots, f(x) > 0 " x ΠR or f(x) < 0 " x ΠR. But
Ê 2ˆ 1
f Ë - ¯ = (4a – 6b + 9c) < 0, therefore
3
9
f(x) < 0 x " x ΠR. fi f(1) < 0
fi a + b + c < 0.
b+c
fi
< -1
a
24. We have 162 – 16c > 0 fi c < 16
If f(x) = 4x2 – 16x + c, then 1 < a < 2 < b < 3
implies f(1) > 0, f(2) < 0, f(3) > 0.
fi c > 12, c < 16, c > 12
Thus, c = 13, 14, 15.
25. If a is a common root of the three equations, then
a2 + aa + 12 = 0, a2 + ba + 15 = 0
and a2 + (a + b)a + 36 = 0
fi ba = – 24, aa = – 21
Thus, a2 – 21 + 12 = 0 fi a = ± 3
As a > 0, we get a = 3.
Therefore, a = – 7, b = – 8.
a
2x
= 2
26.
fi a (x2 – b2) = 2x2
2
x x -b
fi (a – 2)x2 – ab2 = 0
This equation will have no solution if a – 2 and a have
opposite signs, that is , a(a – 2) < 0
fi 0 < a < 2.
a2
4 (a - 2)
27. As the roots are real and distinct,
2
2a + 1
(2a + 1)
>0
fi a2 – 4(a – 2) (2a + 1) > 0
fi
6 - 2 23
6 + 2 23
<a<
7
7
Quadratic Equations 3.45
Write the equation as
a
a-2
=0
x2 –
x+
2a + 1
2a + 1
Note that its roots are a and b.
As a < 1 < b, we must have
a
a-2
+
<0
12 –
2a + 1 2a + 1
fi 2a + 1 – a + a – 2 < 0 fi a < 1/2
1
1
Thus, (6 - 2 23 ) < a <
7
2
2
28. 8x – 10x + 3 = 0 fi (2x – 1) (4x – 3) = 0
fi a, b = 1/2. 3/4
•
\
1
1
 (a n + b n ) = 1 - a + 1 - b = 6
n=0
2
29. b – 4ac = (– a – 2c)2 – 4ac
= a2 + 4c2 > 0
x 2 - bc
, then
2 x - (b + c )
x2 – bc = 2xy – (b + c) y
fi x2 – 2xy + (b + c)y + – bc = 0
As x is real,
y2 – (b + c)y + bc ≥ 0
fi (y – b) (y – c) ≥ 0
fi y cannot lie in (b, c).
31. (a – g ) (a – d ) (b – g ) (b – d )
= (a2 + ra + s) (b2 + rb + s)
= (– pa – q + ra + s) (– pb – q + rb + s)
= [(r – p)a – (q – s)] [(r – p)b – (q – s)]
37. Use 5x + 5–x ≥ 2 " x Œ R.
38. Multiply the first equation by x and subtract from
the second to obtain x = 1.
39.
x+2≥0 fi x≥–2
Now, |(x + 2) (x – 3)| = x + 2
|x + 2| |x – 3| = x + 2
fi x + 2 = 0 or |x – 3| = 1
fi x = – 2, x = 3 ± 1 i.e., x = – 2, 2, 4
40. |x|2 – 3|x| + 2 = 0 fi |x| = 1, 2 fi x = ±1, ±2
41. Use n – 4 > 0 and D < 0
42. Let f (x) = ax2 + bx + c
f (x) < 0 " x Œ R ¤ a < 0, b2 – 4ac < 0.
Now, g(x) = ax2 + bx + c + (2ax + b) + 2a
= ax2 + (2a + b)x + (2a + b + c)
Here a < 0 and (2a + b)2 – 4a(2a + b + c)
= (b2 – 4ac) – 4a2 < 0.
43. Use the fact that expression becomes zero when x
= – 1.
44. Note that p > 0, and
p – 4 < 0, 2p2 – 4 > 0,
and (p – 4)2 – 4(2p2 – 4) ≥ 0
30. Let y =
q–s
˘Èq – s
˘
= (r – p)2 È
ÍÎ r – p – a ˙˚ ÍÎ r – p – b ˙˚
2
È q – sˆ
˘
Ê q – sˆ
= (r – p)2 ÍÊÁ
+ q˙
+ pÁ
˜
˜
Ë r – p¯
ÎË r – p ¯
˚
= (q – s)2 + p(q – s) (r – p) + q(r – p)2
32. Use x2 ≥ 0, |x| ≥ 0
33. x2 – 3|x| + 2 = 0 fi |x| = 1, 2 fi x = ±1 , ±2
34. As p, q ΠR, other root of the equation is 2 Рi 5.
Thus,
p = 4, q = (2 + 5i ) (2 – 5i ) = 9
35. Subtract one equation from the other to obtain
x = 1.
36. Write the expression as
(ap + b)2 + (bp + c)2 + (cp + d)2 £ 0.
fi ap + b = 0, bp + c = 0, cp + d = 0
fi
b c d
= = =– p.
a b c
45.
46.
47.
48.
49.
fi
p < 4, p >
fi
2 < p<
2 and 7p2 + 8p – 32 £ 0.
4
( 15 – 1)
7
f (x)2 + g(x)2 = 0 ¤ f (x) = 0 and g(x) = 0
Use D < 0
Show that D > 0
ax2 + bx + c > 0 " x > 0.
x2 – 8x + p2 – 6p = 0 has real roots
fi 64 – 4(p2 – 6p) ≥ 0
fi p2 – 6p – 16 £ 0 fi (p – 8) (p + 2) £ 0
fi p Π[Р2, 8]
50. –
2a + 3
=–3
a +1
fi
\ Product of roots =
a = 0.
3a + 4
=4
a +1
51. Use that other root of the equation is 3 + 4i to
obtain p = 6, q = 25
52. Let other roots be b, g then
(1 + i) + b + g = 0 and (1 + i)bg = – (1 – i)
fi b + g = – (1 + i), bg = i
1
53. a + b = – p, ab = - 2 .
2p
1
2
2
2
fi a + b = (a + b) – 2ab = p2 + 2
p
1ˆ2
Ê
2
2
fi a + b = Áp– ˜ +2≥2.
Ë
p¯
54. Put
\
x - 1 = t so that x = t2 + 1
x + 3 – 4 x –1 + x +8 – 6 x –1 =1
3.46
Complete Mathematics—JEE Main
fi
t 2 - 4t + 4 + t 2 - 6t + 9 = 1
fi (1 – p)2 – 2(1 + p) (1 – p) + p(p – 1) < 0
and (1 + p)2 – 2(1 + p)2 + p(p – 1) < 0
fi |t – 2| + |t – 3| = 1 fi 2 £ t £ 3
fi2£
fi –1/4 < p < 1 and p > –
x -1 £ 3
fi 4 £ x – 1 £ 9 fi 5 £ x £ 10.
55. |x – x2 – 1| = |2x – 3 – x2| fi |x2 – x + 1| =
|x2 – 2x + 3|
2
3
= |(x – 1)2 + 2|
fi Êx - 1 ˆ +
Ë
4
2¯
fi x2 – x + 1 = x2 – 2x + 3 fi x = 2
c
b
56. sin a + cos a = - , sin a cos a =
a
a
Now 1 = sin2a + cos2a
= (sin a + cos a)2 – 2sin a cos a
fi
1=
fi
2
57. k =
b2
a
2
-
2c
a
fi
a2 = b2 – 2ac
a – b2 + 2ac = 0.
x2 - x + 1
1
3
Ê 1 ˆ
Thus, p Œ Ë- , 1¯
4
b
B
63. a + b = - , a + b + 2h = a
A
2
64. D = 49 + 56(q + 1) > 0.
65. Note that a, b are roots of
b
c
=0
x2 + x +
a
a
As a < – 1, b > 1, we get
b
c
1–
+
<0
a
a
b
c
+
<0
and 1 +
a
a
b c
fi 1+ + < 0
a a
x2 + x + 1
(k – 1)x2 + (k + 1)x + (k – 1) = 0
As x is real (k + 1)2 – 4(k – 1)2 ≥ 0
fi (k + 1 + 2k – 2) (k + 1 – 2k + 2) ≥ 0
fi (3k – 1) (3 – k) ≥ 0 fi 1/3 £ k £ 3.
a2 + ba + ca = 0 and a2 + ca + ab = 0
fi (b – c)a + a(c – b) = 0 fi a = a.
Now, show that a + b + c = 0 and use the fact that
other roots are b and c.
mx2 + 3x + 4 < 5 (x2 + 2x + 2)
fi (m Р5)x2 Р7x Р6 < 0 " x ΠR
fi m – 5 < 0 and (– 7)2 + 24(m – 5) < 0
fi m < 5 and 24m – 71 < 0 fi m < 71/24
Use the fact that expression becomes equal to 1 for
x = a, b, c.
As ax2 + bx + c has no real zeros.
ax2 + bx + c > 0 " x ΠR or ax2 + bx + c < 0 "
x ΠR.
Since a(– 1)2 + b(– 1) + c < 0, we get ax2 + bx +
c<0"xŒR
Thus, a < 0 and c < 0 fi ac > 0.
x2 – 2x – p2 + 1 = 0
fi (x – 1)2 = p2
fix=1±p
Let a = 1 – p, b = 1 + p
Let f(x) = x2 – 2(p + 1)x + p(p – 1)
As f(x) = 0 has distinet real roots,
(p + 1)2 – p(p – 1) > 0
1
fi 3p + 1 > 0 fi p > –
3
Also, f(1 – p) < 0,
and f(1 + p) < 0
fi
58.
59.
60.
61.
62.
Fig. 3.22
66. Let f (x) = ax2 + bx + c.
Now, (a + c)2 < b2
¤ (a – b + c) (a + b + c) < 0
fi f (– 1) f (1) < 0.
Thus, ax2 + bx + c = 0 has a real root between – 1
and 1.
The other root also must be real.
67. If x is an integer and
(x – a) (x – 10) = – 1
fi x – 10 = 1 and x – a = – 1 or x – 10 = – 1 and
x–a=1
fi x = 11 and a = x + 1 = 12 or x = 9 and a =
x–1=8
x
2 x
3
68. 41.5 Ê ˆ + 90.5 Ê ˆ = 10
Ë 3¯
Ë2 ¯
Put Ê 3 ˆ
Ë2 ¯
x
= t, so that
3t2 – 10t + 8 = 0
4
fi (3t – 4) (t – 2) = 0 fi t = , 2
3
x
4
3ˆ
Ê
fi
= ,2
Ë2 ¯
3
3t + 8/t = 10
fi
Quadratic Equations 3.47
fix=
log ( 4 / 3)
log 2
,
log (3 / 2 ) log (3 / 2 )
2
69. Put 2sin x = t, so that equation becomes
5¥2
t+
= 7 fi t2 – 7t + 10 = 0 fi t = 2, 5
t
But 1 £ t £ 2. Thus, sin2x = 1
70. Let the two distinct roots lying between 0 and 1 be
a, b such that a < b.
As f (x) = x2 – 2x + k is a differentiable and
f (a) = f (b) = 0, therefore by the Roll’s theorem
there exists g Œ (a, b) à (0, 1) such that f¢ (g) = 0
fi 2g – 2 = 0 fi g = 1. Not possible.
71.
a
a +1
a
a +1 c
b
+
.
=
= - , and
a -1
a
a -1 a
a
a
b c 2
Now, (a + b + c)2 = a 2 Ê 1 + + ˆ
Ë
a a¯
a + 1 ¸ a + 1˘
È Ï a
+
= a 2 Í1 - Ì
˝+
1
a
a ˛ a - 1 ˙˚
Î Ó
77. y =
2
fi
fi
È a
a + 1ˆ 2
a a + 1˘
= a ÍÊÁ
+
-4
.
˙
˜
a ¯
a -1 a ˚
ÎË a - 1
2
È 2
˘
= a 2 Í b - 4c ˙ = b2 - 4 ac
2
a˚
Îa
72. Let x = 3m + r, 0 £ r £ 2.
x2 + x + 1 = (9m2 + 6mr + 3m) + (r2 + r + 1)
Note that x2 + x + 1 will be divisible by 3 if and
only if r = 1.
=
=
1
(a – b )2 – x2 – (a + b )x + ab
4
(a - b )2 + 4ab
4
{
- x-
{ }
1
a
(a + b )2 - x +
2
2
8 + 2 8 + 2 8 + 2 8 +
y2 = 8 + 2y fi y2 – 2y – 8 = 0
y = 4, – 2. As y > 0, y = 4
Ê px ˆ
2
=
78. sin Á
+1≥1
Ë 2 3 ˜¯ ( x – 3 )
a
a + 1ˆ 2
= a 2 ÊÁ
Ëa -1
a ˜¯
73. f (x) =
are not sufficient to guarantee that the roots are
real. For instance, x2 + 0x + 0.5 = 0 does not have
real roots.
75. Let f (x) = ax2 + bx + c.
Note that f (0) = c > 0,
f (– 1) = a – b + c < 0 and
f (1) = a + b + c < 0
Thus, y = ax2 + bx + c
represents a parabola that
Fig. 3.24
open downwards. See
Fig.3.21
Thus [a] = – 1 and [b] = 0
fi [a] + [b] = – 1
76. x2 – 2mx + m2 – 1 = 0 fi (x – m)2 = 1
fix–m=±1 fi x=m±1
Therefore, – 2 < m – 1 < m + 1 < 4
fi – 1 < m < 3.
a +b
2
}
2
+
(a + b )2
4
2
2
£
a
2
a2
Thus, maximum value f (x) is
which is
2
a
attained when x = –
2
74. Let a, b be roots of f(x) =
x2 + bx + c, then 0 < a, b
< 1 fi 0 < a + b < 2 and
ab < 1
fi – b < 2 and c < 1
fi b > – 2 and c < 1
However, these conditions
Ê px ˆ
But sin Á
£ 1. Thus, sin ÊÁ p x ˆ˜ = 1
Ë 2 3 ˜¯
Ë 2 3¯
Therefore, x = 3
79. 2(3 – x) = |x + 2| ≥ 0 fi x £ 3
If x £ – 2, the equation becomes
2(3 – x) = – x – 2 fi x = 8. Not possible
If – 2 < x £ 3, the equation becomes
2(3 – x) = x + 2 fi 3x = 4 fi x = 4/3.
80. Use product of roots < 0
81. Put x2 – 5x = y;
so that y2 – 7y + 6 = 0 fi y = 1, 6
\ x2 – 5x = 1 or x2 – 5x = 6
fi x2 – 5x – 1 = 0 or x2 – 5x – 6 = 0
But x2 – 5x – 6 = 0 has integral roots and roots of
x2 – 5x – 1 = 0 are
1
x = (5 ± 29 )
2
2
Ê x - 1ˆ
82. Put Á
= y to obtain
Ë x + 1˜¯
y2 – 13y + 36 = 0 fi y = 4, 9
x -1
= ± 2, ± 3
\
x +1
Thus, the given equation has four real roots.
83. 9x + 2 – 6(3x + 1) + 1 = 0
Put 3x + 1 = t so that t2 – 6t + 1 = 0
Fig. 3.23
fit=
6 ± 32
=3±2 2
2
3.48
Complete Mathematics—JEE Main
fi x + 1 = log3 (3 + 2 2 ) > 1
or x + 1 = log3 (3 – 2 2 ) < 0
2
Ê 2 x – 5ˆ
= t, so that equation becomes
84. Put Á
Ë 3 x + 1 ˜¯
81t2 – 45t + 4 = 0
fi 81t2 – 36t – 9t + 4 = 0
fi (9t – 1) (9t – 4) = 0
2x – 5
1 2
1 4
=± ,±
fit= , fi
3x + 1
3 3
9 9
Thus, the given equation has four rational roots.
85. Put x2 + 3x + 2 = t to obtain t2 – 8(t – 2) – 4 = 0
fi t2 – 8t + 12 = 0 fi (t – 2) (t – 6) = 0
fi t = 2, 6
But both x2 + 3x + 2 = 2
and x2 + 3x + 2 = 6 have integral roots.
x
1
5
= t to obtain t + =
x-3
t
2
1
fi t = 2,
2
x
x
1
1
\
= 2,
fi
= 4,
x
3
2
4
x-3
89.
90.
86. Put
87. Put x + 1/x = t to obtain
4{t2 – 4} + 8t = 29 or 4t2 + 8t – 45 =
45
45
fi t2 + 2t =
fi (t + 1)2 =
+1=
4
4
7
9 5
fi t = -1 ± = - ,
2
2 2
Thus,
x+
91.
92.
0
2
Ê7 ˆ
Ë2 ¯
1
9 5
= - ,
x
2 2
9
5
x + 1 = 0, x 2 - x + 1 = 0
2
2
The first equation has irrational roots and the
second equation has rational roots.
88. Divide the given equation by x2 to obtain
fi
x2 +
fit=1± 3.
But t ≥ 0, therefore t = 1 + 3
1
fi 2x + 1 = 1 + 2 + 2 3 fi x = (3 + 2 3 )
2
93. Note that – 13 £ x £ 13 and
13 – x2 = x2 + 10x + 25
fi x2 + 5x + 6 = 0 fi x = – 2, – 3.
94.
fix=2
x-3
fi 3x2 – 6x – 25 = 0
28
28
fi x = 1±
fi (x – 1)2 =
3
3
1
1
,= - –4
x
2
Note that 1 –
2
x +1
1 x +1
= - ,
=–4
x
x
2
fi 2x2 + x + 2 = 0,
x+2 - x-2 =
x-3
fi x + 3 = 2 x 2 - 4 fi x2 + 6x + 9 = 4(x2 – 4)
1
fit= - ,–4
2
fi
or
x-2
fi x + 2 + x – 2 – 2 x2 - 4 = x – 3
1
Put x + = t to get 2(t2 – 2) + 9t + 8 = 0
x
2
x 2 - 4 - ( x - 2) = x 2 - 5x + 6
(1)
is defined for x = 2 or x £ – 2 or x ≥ 3. Write
(1) as
x - 2 [ x + 2 - x - 2] =
1
1
2 ÊÁ x 2 + 2 ˆ˜ + 9 Ê x + ˆ + 8 = 0
Ë
¯
Ë
x¯
x
\ x+
The first equation has non-real complex roots
and the second equation has irrational roots. viz.
–2± 3.
1
Put x - = t to obtain 4t2 – 4t + 1 = 0
x
1 1
fi t= ,
2 2
1
x2 - 1
1
1
Now, x - =
fi
=
2
2
x
x
2
fi 2x – x – 2 = 0
1 1
\ Sum of roots of the given equation = + = 1.
2 2
(3x – 2)(x – 1) (3x + 1)(x – 2) = 21
fi (3x2 – 5x + 2) (3x2 – 5x – 2) = 21
fi (3x2 – 5x)2 – 4 = 21 fi 3x2 – 5x = ± 5
fi 3x2 – 5x – 5 = 0 or 3x2 – 5x + 5 = 0
The first equation has irrational roots and the
second has imaginary roots.
Note that x ≥ 2 and
fi x – 2 = 3 x - 6 fi x2 – 4x + 4 = 3x – 6
fi x2 – 7x + 10 = 0 fi x = 2, 5
Both of these satisfy the given equation.
1
Note that x ≥ - . Put 2 x + 1 = t, so that
2
2t = t2 – 1 – 1 or t 2 – 2t + 1 = 3
x2 + 4x + 1 = 0
95.
28
< – 2 and 1 +
3
28
>3
3
x 2 - 4 x + 3 + x 2 - 7 x + 12 = 3 x - 3
is defined for x = 3 or x ≥ 4.
(1)
Quadratic Equations 3.49
Write (1) as
x - 3 [ x - 1 + x - 4] = 3 x - 3
fi x = 3 or
x -1 + x - 4 = 3
fi x = 3 or
x -1 = 3 - x - 4
fi x = 3 or x – 1 = 9 - 6 x - 4 + x - 4
fix = 3 or 6 x - 4 = 6 fi x = 3 or x = 5.
Thus, product of roots = 15.
96. Note that 2b2 + 77b + 18 = 0
fi 2 + 77/b + 18(1/b)2 = 0
Thus, a and 1/b are roots of the equation
18x2 + 77x + 2 = 0
a 2 1
\
a + 1/b = – 77/18 and =
=
b 18 9
1 a
77 1
ab + a + 1
25
Now,
= a+ + =+ =b
b b
18 9
6
97. b2 = 7b – 8.
16
16
Thus,
+ 3b 2 - 19b =
+ 3(7b - 8) - 19b
a
a
16
+ 2b - 24
=
a
Also, a =
\
Thus
7 + 17
7 - 17
,b=
2
2
(
)
2 7 - 17
1
2
7 - 17
=
=
=
a
49 - 17
16
7 + 17
16
+ 3b 2 - 19b
a
(
) (
= a2a2 – 2a2a2 < 0
and a2b2 + 2bb + 2c
= a2b2 + 2a2b2 > 0
As a2x2 + 2bx +2c is continuous on [a, b], by statement-2 there exists g such that a2g 2 + 2bg + 2c = 0
102. As y = f(x) open upwards, a lies between the roots of
f(x) = 0, if and only if f(a) < 0. Therefore, statement-2
is true.
If f(x) < 0 for 1 £ x £ 2, then 1 and 2 lie between roots
of f(x) = 0. Thus, we must have
4p2 – 4p > 0 fi p(p – 1) > 0
fi p < 0 or p > 1.
Also, f(1) = 1 + 2p + p < 0 fi p < – 1/3 and
f(2) = 4 + 4p + p < 0 fi p < – 4/5.
\ p Œ (– •, –4/5)
103. Statement-2 is false as x2 + 5x + 4 = 0 meets both the
conditions but does not have positive roots.
As LHS ≥ 0, 1 – 2x ≥ 0 fi 2x £ 1 fi x £ 0.
Squaring, we get
a(2x – 2) + 1 = 1 – 2(2x ) + 22x
fi y2 – (a + 2)y + 2a = 0
fi (y – a) (y – 2) = 0 fi y = 2, a.
y = 2x = 2 fi x = 1. Not possible.
Thus, y = a. This implies
0 < 2x = a £ 1
Therefore 0 < a £ 1, and x = log2a
104. Put t = x +
fi
)
= 7 - 17 + 7 - 17 - 24 = – 10
98. As a is a root of x4 + x2 – 1 = 0, we get
a4 + a2 – 1 = 0 fi a6 + a4 – a2 = 0
fi a 6 + a 4 – (1 – a 4) = 0 fi a 6 + 2a 2 = 1
\ (a 6 + 2a 2)2012 = 1
99. Use the fact that sum and product of the roots of
x2 – x – k = 0 are 1 and – k respectively.
100. Suppose a2(a + k) = b2(b + k) = c2(c + k) = t, then
a, b, c are roots of x3 + kx2 – t = 0 thus, bc + ca
+ ab = 0 and abc = t
1 1 1 bc + ca + ab
fi + + =
=0
a b c
abc
101 For truth of Statement-2, see theory
f(a) = a2a2 + ba + c = 0 and
g(b ) = a2b 2 – bb – c = 0
We have a2a2 + 2ba + 2c
1
=
t
\t–
x 2 + b2
1
x 2 + b2 + x
=
x 2 + b2 - x
b2
b2
= 2x
t
Thus,
b2
)t = b2 + 2at – t2
t
= a2 + b2 – (t – a)2
Therefore, maximum value of f(x) is a2 + b2 which is
f(x) = (2a – t +
b2 ˆ
1Ê
ÁË a - ˜¯
a
2
Statement-2 is false as maximum value of f(x) = ax2
+ bx + c, a < 0 is (4ac – b2)/4a.
105. For truth of statement-2 see theory
Let P(x) = ax2 + bx + c.
As P(x) = x does not have real roots, P(x) > x " x Œ
R or P(x) < x " x ΠR.
attained when t = a or x =
3.50
Complete Mathematics—JEE Main
Suppose P(x) < x " x ΠR.
fi P(P(x)) < P(x) < x ⁄ x Œ R
fi a(ax2 + bx + c)2 + b(ax2 + bx + c) + c – x = 0
cannot have real roots.
106. Statement-2 is true as |a + b| = |a| + |b|
¤ a, b are both non-negative or both non-positive.
As ax2 + bx + c = 0 has imaginary roots, ax2 + bx + c
> 0 " x ΠR or ax2 + bx + c < 0 " x ΠR.
Since a(–1)2 + b(–1) + c = a – b + c < 0,
we get ax2 + bx + c < 0 " x ΠR.
Now, |ax2 + bx + c| + |x + y|
= a Ê x 2 + y ˆ + (b + 1) x + c
Ë
a¯
2
= |(ax + bx + c) + (x + y)|
¤ (ax2 + bx + c) (x + y) ≥ 0
¤ x + y £ 0 as ax2 + bx + c < 0 " x Œ R.
107. Statement-2 is true. See Theory
Put x + c = t, so that
[ t + ( a - c )] [ t + ( b - c )]
f(x) =
t
( a - c ) (b - c )
=t+
+ [(a – c) + (b – c)]
t
2
È
2
( a - c ) (b - c ) ˘
= Í t˙ + [ a - c + b- c]
Î
˚
t
Thus, minimum value of f(x) is
and it is attained when t =
x=–c+
(
2
a - c + b - c)
1
≥ 2.
y
112.
113.
114.
115.
1
È
tan 2 a ˘ 2
x +x+
≥ 2 Í x2 + x
˙
ÍÎ
x2 + x
x 2 + x ˙˚
= 2 tan a
(a – b)2 + 4(a + b – 1) ≥ 0 " b ŒR.
fi b2 – 2b(a – 2) + a2 + 4a – 4 ≥ 0 " b ŒR.
fi 4(a – 2)2 – 4(a2 + 4a – 4) < 0
fi – 8a + 8 < 0 fi a > l
\ a Œ (1, •)
a + b = –b < 0 and ab = c < 0.
As ab < 0, and a < b,
a < 0 < b < – a = |a|
As x2 + (a + b) x + c = 0 has no real roots,
and the coefficient of x2 = 1 > 0,
f (x) = x2 + (a + b) x + c > 0 x ŒR.
fi (a + b)2 – 4c < 0.
Also f (0) = c > 0, f (1) = a + b + c + 1 > 0
fi c (a + b + c) + c > 0
Next, f (0) f (–1) > 0
fi c (1 – a – b + c) > 0
fi c – c(a + b – c) > 0
As x – 1 < [x] £ x, we get –x £ – [x] < – x + 1
= x2 – 2x – 2 £ x2 – 2[x] – 2 < x2 – 2x
= (x – 1)2 –3 £ x2 –2[x]–2 < (x – 1)2 – 1
2
(a - c ) (b - c ) or
108. If a π p, f(x) – g(x) = 0 is a quadratic equation whose
coefficient of x2 π 0.
\ f(x) – g(x) = 0 has two real roots
Thus, statement-2 is true
Let h(x) = f(x) – g(x)
= (a – b) x2 + (b – q) x + (c – r)
If h(x) = f(x) – g(x) = 0
for three distinct real values of x, then a – p = 0, b – q
= 0, c – r = 0
fi a = p, b = q, c = r.
\ Statement-1 is also true.
However, statement-2 is not a correct explanation for
Statement-1.
109. As A.M. ≥ G.M, we get
fiy+
Level 2
111. As A.M ≥ G.M, we get
(a - c ) (b - c ) > – c.
1Ê
1ˆ
Ê 1ˆ
y + ˜ ≥ yÁ ˜ = 1
Á
Ë y¯
y¯
2Ë
Thus, statement-2 is true.
Since, sin2 x = 0,
does not satisty sin4 x + p sin2 x + 1 = 0, we get
0 < sin2 x £ 1, for all x satisfying sin4 x + p sin2 x + 1 = 0
For such a value of sin x,
1
– p = sin2x +
≥2
sin 2 x
fi p £ – 2 or p Œ (– •, –2]
Let f(t) = t 2 + pt + 1
As f(0) = 1 > 0 and f(1) = 1 – p + 1 £ 0,
we get there is at least one value of t Π(0, 1] such that
f(t) = 0.
110. As b2 Р4ac < 0, f(x) > 0 " x ΠR or f(x) < 0 " x ΠR.
As f(0) = c > 0, f(x) > 0 " x ΠR.
Thus, statement-2 is true.
As f(x) = 0 does not have real roots, b2 – 4ac < 0. By
the statement-2, f(x) > 0 " x ΠR.
fi f(–1) = a –b + c > 0 fi a > b – c.
Also b2 – 4ac < 0
fi a > b2/4c
¸
Ï b2
Hence, a > max Ì , b - c ˝ .
Ó 4c
˛
tan 2 a
Quadratic Equations 3.51
Note that (x – 1)2 – 3 £ 0 for – 3 + 1 £ x £ 3 + 1
and (x – 1)2 – 1 > 0 for x < 0 or x > 2,
The roots of x2 – 2[x] – 2 = 0
(
If x Π( -
)
3 + 1, 0) , [x] = –1,
lie in - 3 + 1, 0 or (2, 1 +
3)
and x2 – 2[x] – 2 = 0 fi x2 = 0 or x = 0.
This is not possible.
Suppose x Π(2, 1 + 3 ), then [x] = 2,
x2 – 2[x] – 2 = 0 fi x2 = 6 or x =
6.
Note that x = 6 satisfies x2 – 2[x] – 2 = 0
116. If [x] is odd, then
[x]2 + a[x] + b is odd and if [x] is even, then [x]2
+ a[x] + b is also odd
Thus, [x]2 + a[x] + b can never take value 0.
117. Let a = sina, b = sinb, c = sing .
Note that a, b, c are distinct and 0 < a, b, c < l.
We can write the given equation as
f (x) = (x – b) (x – c) + (x – c) (x – a)
+ (x – a)(x – b) = 0.
Assume that a < b < c.
Note that f (a) = (a – b) (a – c) > 0
f (b) = (b – c) (b – a) < 0
and
f (c) = (c – a) (c – b) > 0.
Thus, f (x) = 0 has a root in (a, b) and a root in (b, c).
118. Let f (x) = x2 + (2a + a) x + a 2 + aa + b.
f (0) = a2 + aa + b = 0
Let other root of f (x) = 0 be r, then
0 + r = –(2a + a) = –2a + a + b = b – a.
Thus, two roots are 0 and b – a.
119. As roots are real
(a + b + c)2 – 3l (ab + bc + ca) ≥ 0
(a + b + c)2
= 3l £
ab + bc + ca
Since, a, b, c are sides of triangle.
b + c – a > 0, c + a – b > 0, a + b – c > 0
fi 2(ab + bc + ca) > a2 + b2 + c2
fi (a + b + c)2 < 4 (ab + bc + ca)
Thus, 3l < 4 fi lΠ(Х, 4/3).
120. tan q + cot q = 2a and tanq cotq = b
Now, 2a = tanq + cotq ≥ 2 if tanq > 0
and 2a = tanq + cot q £ – 2 if tanq < 0.
fi 2|a| ≥ 2 fi |a| ≥ 1
Thus, least value of |a| is 1.
121. Let f (x) = x3 + x2 – 5x – 1, then
f (–3) < 0, f (–2) > 0, f (–1)> 0, f (0) < 0,
f (1) < 0, f (2) > 0.
Thus, a root of f (x) = 0 lies in each of the three intervals:
(–3, –2), (–1, 0),(1, 2).
\ [a] + [ b ] + [g ] = –3 + (–1) + 1 = –3
122. tan A tan B = p, tan A tan B = q.
tan A + tan B
p
=
, we get
As, tan (A + B) =
1 - tan A tan B
1- q
sin2 (A + B) =
=
tan 2 ( A + B)
1 + tan 2 ( A + B)
p2
p 2 /(1 - q)2
2
2 =
1 + p /(1 - q)
(1 - q )2 + p 2
123. Put x - 1 = t fi x – 1 = t2, so that equation
becomes;
t 2 + 1 + 3 + 4t + t 2 + 1 + 8 + 6t = 1
fi |t + 2| + |t + 3| = 1
Not possible as t ≥ 0.
Thus, the equation has no solution.
124. Put x 2 + 11 = t fi x2 + 11 = t2, so that equation
becomes
t 2 + t - 11 + t 2 - t - 11 = 4.
Since (t2 + t – 11) – (t2 – t – 11) ∫ 2t,
we get
t
t 2 + t - 11 - t 2 - t - 11 =
2
t
\ 2 t 2 + t - 11 = 4 +
2
1
fi 4(t2 + t – 11) = 16 + t 2 + 4t fi t2 = 16
4
fi x2 +11 = 16 fi x = ± 5
125. Note that a, b, c are roots of
x
y
z
+
+
=1
t t- p t-q
¤ t3 – (x + y + z + p + q) t2
+[(p + q) x + qy + pz] t – pqx = 0.
\ product of roots = pqx
abc
pq
126. Putting x = 0, 1, 1/2, we get
–1 £ c £ 1 , – 1 £ a + b + c £ 1,
fi abc = pqx fi x =
-1 £
1
1
a + b + c £1,
4
2
3.52
Complete Mathematics—JEE Main
fi – 4 £ – a – 2b – 4c £ 4
As – 4 £ 4a + 4b + 4c £ 4, we get
– 8 £ 3a + 2b £ 8.
Also, – 8 £ a + 2b £ 8
fi –16 £ 2a £ 16 fi |a| £ 8
127. p + q and p + r are roots of
at2 + 2bp (t – p) + c = 0
or at2 +2bpt + c – 2bp2 = 0.
We have
|q – r|2 = |(p + q) – ( p + r)|2
= [( p + q) + ( p + r)]2 – 4( p + q)( p + r)
(
2
4 c - 2bp 2
Ê -2bp ˆ
= Á
Ë a ˜¯
a
)
4
= 2 ÈÎb 2 p 2 - ac + 2abp 2 ˘˚
a
4
= 2 [(2a + b) bp2 – ac]
a
fi |q – r|
2
(2a + b)bp 2 - ac
=
|a|
128. Put x2 = y. The given equation will have four real
roots if
f ( y) = ay2 + by + c = 0
has two non-negative roots.
This is possible if
b
– ≥ 0 , af (0) ≥ 0, b2 – 4ac ≥ 0
a
fi ab £ 0, ac ≥ 0.
This condition is met if a > 0, b < 0, c > 0 or a < 0, b
> 0, c < 0
129. As x = 1 satisfies
(b – c) x2 + (c – a) x + (a – b ) = 0,
x = 1 must satisfy x2 + mx + 1 = 0
fi 1 + m + 1 = 0 fi m = –2.
130. Eliminating x from x + y + z = 4,
and x2 + y2 + z2 = 6, we get
y2 + z2 + (4 – y – z)2 = 6
fi 2y2 – 2(4 – z)y + (4 – z)2 + z2 – 6 = 0
As y is real, we get
(4 – z)2 –2[(4 – z)2 + z2 – 6] ≥ 0
fi 3z2 – 8z + 4 £ 0
fi (3z – 2) (z – 2) £ 0
fi 2/3 £ z £ 2.
Thus, maximum value of z is 2.
Previous Years’ AIEEE/JEE Main Questions
1. a, b are root of x2 = 5x – 3 or x2 – 5x + 3 = 0
fi a + b = 5, ab = 3.
2
2
2
Now, a + b = a + b = (a + b ) - 2ab
b a
ab
ab
=
25 - 6 19
=
3
3
Êaˆ b
and Á ˜ Ê ˆ = 1
Ë b¯ Ëa¯
Thus, equation whose roots are a/b and b/a is
x2 – (19/3)x + 1 = 0
3x2 – 19x + 3 = 0
or
2. |a – b| = |a1– b1|
fi (a – b)2 = (a1 - b1)2
fi (a + b)2 - 4ab = (a1 + b1)2 - 4a1b1
fi a2 - 4b = b2 – 4a
fi (a – b)(a + b + 4) = 0 fi a + b + 4 = 0
[ a π b]
1 3 1 2
ax + bx + cx.
3
2
As f
f is continuous on [0, 1] and
f is differentiable on (0, 1). Also
3. Let f(x) =
f(0) = 0 and f(1) =
1
(2a + 3b + 6c) = 0
6
a Π(0, 1)
such that f ¢(a) = 0.
Thus, ax2 + bx + c = 0 has at least one root in
[0, 1].
4. For each x ΠR, t ΠR
t2x2 + |x| + 9 ≥ 9 > 0
Thus, t2x2 + |x| + 9 = 0 does not have real roots.
5. Let a and 2a be the roots of the given equation,
then
(a2 – 5a + 3)a2 + (3a – 1)a + 2 = 0
(1)
(a2 – 5a + 3)(4a2) + (3a – 1)(2a) + 2 = 0
(2)
get
(3a – 1) (2a) + 6 = 0
Quadratic Equations 3.53
a π 1/3. Therefore,
a = – 3/(3a – 1)
Putting this value in (1) we get
(a2 – 5a + 3)(9) – (3a –1)2(3) + 2(3a – 1)2 = 0
2
2
fi
9a – 45a + 27 – (9a – 6a + 1) = 0
fi
–39a + 26 = 0
fi
a = 2/3.
For a = 2/3, the equation becomes x2 + 9x + 18 =
0, whose roots are –3, –6.
6. Let a, b be the roots of ax2 + bx + c = 0
1
1
+ 2
2
a
b
a + b=
a = –1 and – (1– p) = 1– p
fi
a = 1 and 1 – p = 0
Thus, equation is x2 + x = 0
fi
x = 0, –1
10. Let a be the other root of x2 + px + 12 = 0
\
(a)(4) = 12 and a + 4 = p
fi
a = 3 and 3 + 4 = p
As x2 + px + q = 0 has equal roots, we must have
p2 – 4q = 0
a 2 + b 2 (a + b ) 2 - 2ab
=
(ab ) 2
(ab ) 2
=
( - b / a ) 2 - 2( c / a )
-b
=
a
(c / a ) 2
2
2
b c
b
2c
- ◊ 2= 2 a a
a
a
fi
fi
fi
Thus, p = 7,
We are given
fi
a + (1 – p) = – p and a (1– p) = 1 – p
1 2 49
p =
4
4
11. See solution to Question No. 3
fi q=
12. We have
Pˆ
Ê Q ˆ -b and tan ÊÁ P ˆ˜ tan ÊÁ Q ˆ˜ = c
tan ÊÁ ˜ + tan Á ˜ =
Ë 2¯ a
Ë 2¯
Ë 2¯ a
Ë 2¯
2a2c = ab2 + bc2
Also
abc to obtain
fi
a c b
= +
b a c
c a b
, , are in A.P.
a b c
a b c
, , are in H.P.
c a b
x2 – 3|x| + 2 = 0
P + Q = p – R = p/2
P Q p
+ =
2 2 4
P Q
p
tan Ê + ˆ = tan Ê ˆ
Ë 2 2¯
Ë 4¯
2
fi
fi
7.
fi
tan( P / 2) + tan(Q / 2)
=1
1 - tan( P / 2) tan(Q / 2)
-bla
-b
=1fi
=1
1 - cla
a-c
fi
fi
2
fi
|x| – 3|x| + 2 = 0
fi
(|x| – 1) (|x| – 2) = 0
fi
|x| = 1, 2
fi
x = ± 1, ± 2
8. Let two numbers be a and b. We are given
fi
–b = a – c
or
c=a+b
13. Let a, b be roots of x2 – (a – 2)x – a – 1 = 0
Then
a + b = a – 2, ab = – (a + 1)
Now,
a2 + b2 = (a + b)2 - 2ab
= (a – 2)2 + 2(a + 1)
a+b
= 9, ab = 4
2
fi a + b = 18 and ab = 16
= a2 – 2a + 1 + 3
\
= (a – 1)2 + 3
x2 – (a + b)x + ab = 0
or
= a2 – 4a + 4 + 2a + 2
Thus, a2 + b2 is least when a = 1
2
x – 18x + 16 = 0
9. Let other root be a. Then
14. We know that if a and b are roots of x2 – bx + c
= 0, then (a – b)2 = discriminant = b2 – 4c.
3.54
Complete Mathematics—JEE Main
In the present case b = a + 1.
fi
1 = b2 – 4c
\
3(y – 1)x2 + 9(y – 1)x + 7y – 17 = 0
As x is real
81(y – 1)2 – 12(y – 1) (7y – 17) ≥ 0
15. We can write the given equation as
(x – k)2 = 5 – k.
Thus, for roots to be real we must have 5 – k ≥ 0
or k £ 5.
Also, for k £ 5,
x = k ± 5-k
For both the roots to be less than 5, we must have
that larger of the two roots is less than 5. Therefore.
fi
3(y – 1) [27(y – 1) – 4(7y – 17)] ≥ 0
fi
3(y – 1) (41 – y) ≥ 0
fi
(y – 1)(y – 41 ) £ 0
fi
1 £ y £ 41
20. a + b = – a, ab = 1
|a – b| <
5
2
(1)
fi
|a – b| < 5
This does not hold for k = 5. For k < 5, we can
write (1) as
fi
(a + b)2 - 4ab < 5
fi
a2 – 4 < 5
fi
a2 < 9
fi
–3 < a < 3
fi
a Π(- 3, 3)
fi
k + 5-k <
1< 5 - k
\
or
fi5 5-k < 5-k
5–k>1
or
k<4
k Œ (– •, 4).
16. Let f (x) = an xn + an – 1xn –1 + ... + a1x. Then, f(0)
= 0 and f(a) = 0.
$ b Œ(0, a) such that f ¢(b)
= 0.
Thus,
nan xn – 1 + (n – 1)an –1xn – 2 + ... + a1 = 0
has a positive root b which is smaller than a.
1
17. Let a = tan 30°, b = tan 15°, then a =
3
and b = 2 - 3
1
+ (2 - 3 )
\ -p =
3
2
1
(2 - 3 ) =
-1
and q =
3
3
Thus,
\
2
3
q- p=
+ 2 - 3 -1 = 1
3
2+q–p=3
21. Let roots of x2 – 6x + a = 0 be a, 4m and that of
x2 – cx + 6 = 0 be a, 3m where m is an integer
We have,
a + 4m = 6, a (4m) = a
and
fi
a + 3m = c, a (3m) = 6
am = 2 fi
a=8
8
= 6 fi a2 – 6a + 8 = 0
a
(a – 2) (a – 4) = 0 fi a = 2, 4.
Also, a +
fi
As m is an integer, a π 4 (
\
am = 2)
a=2
22. As roots of bx2 + cx + a
4ab < 0. Now,
3b2x2 + 6bcx + 2c2
2
18. x – 2mx + m – 1 = 0
fi
(x – m)2 = 1
fix=m±1
As both the roots lie between –2 and 4, we get
2
= 3 ÈÍb 2 x 2 + 2bcx + c 2 ˘˙
3 ˚
Î
–2 < m ± 1 < 4
1
= 3 ÈÍ(bx + c)2 - c 2 ˘˙
3 ˚
Î
fi
= 3 (bx + c)2 – c2 ≥ - c2 > – 4ab
–1 < m < 3.
3x 2 + 9 x + 17
19. Let y =
3x 2 + 9 x + 7
\ 3b2x2 + 6bcx + 2c2 > Р4ab " x ΠR.
23. x2 – x + 1 = 0
c2 –
Quadratic Equations 3.55
fi (p – 5) (p + 2) = 0
1
3
i
x = – w, – w2 where w = - +
2 2
Let a = -w and b = – w2
fi
We have a
2009
+b
= (– w)2009 + (– w2)2009
= (– 1) [w2007 w2 + (w2)2007 w4]
= (– 1) (w2 + w) = (– 1) (– 1) = 1
24. As a π a1, p(x) = f(x) – g(x
x = –1, we get p(x)
nomial. As p(x
must be of the form p(x) = k(x + 1)2 where k =
a – a 1.
p(–2) = 2, we get
As
2 = k(– 2 + 1)2 fi
Thus,
p(x) = 2(x + 1)2
\
p(2) = 2(2 + 1)2 = 18
Thus, p lies in the set {–2, 5}
28.
2009
k = 2.
p = – 2, 5
a 2 b 2 a3 + b 3 - p
+
=
=
q
b
a
ab
Ê a2 ˆ Ê b2 ˆ
and Á ˜ Á ˜ = ab = q
Ë b ¯Ë a ¯
Thus, required equation is
Ê pˆ
x2 - Á - ˜ x + q = 0
Ë q¯
qx2 + px + q2 = 0
or
29. As the roots are real and distinct
(a + 1)2 – 4(a2 + a – 8) > 0
3a2 + 2a – 33 < 0
fi
(3a + 11)(a – 3) < 0 fi -
11
<a<3
3
(1)
25. We can write the given equation as
y – 1/y – 4 = 0
where y = esinx
fi y2 – 4y – 1 = 0
fi y=
4 ± 16 + 4
=2± 5
2
As y > 0, y = 2 + 5
fi sin x = ln( 2 + 5 ) > ln(4) > 1
This is not possible.
Thus, equation has no real roots.
26. x2 + 2x + 3 = 0
fi
fi
Also, for
22 – 2(a + 1) + a2 + a – 8 < 0
fi
a2 – a – 6 < 0 fi (a – 3) (a + 2) < 0
fi
–2<a<3
(2)
From (1) and (2), we get – 2 < a < 3
(x + 1)2 = 2i2
30. 0 <
x = -1 ± 2i
x-5
x-5
=
(
7)( x - 2)
x
+
x + 5 x - 14
2
Sign of E in different intervals are shown below
As a, b, c ΠR, if
-1 ± 2i (or -1 - 2i ) is a root of ax2 + bx + c
= 0, the other root must be -1 - 2i (or -1 ± 2i )
Thus, x2 + 2x + 3 = 0 and ax2 + bx + c = 0 have
both roots in common, therefore
a b c
= =
1 2 3
fi a:b:c=1:2:3
2
27. ( 10 ) = |a – b|2 = (a + b)2 – 4ab
fi 10 = p2 – 3p
fi p2 – 3p – 10 = 0
Thus,
fi
x-5
>0
( x + 7)( x - 2)
–7 < x < 2 or x > 5.
Therefore the least integral value a of x is – 6.
a2 + 5a – 6
This value of a
= 0.
3.56
Complete Mathematics—JEE Main
31. 4 =
fi
fi (k2 – 4)2 = 0 fi k = 2
a + b -q / p -q
=
=
ab
r/ p
r
q + 4r = 0
Also,
Thus,
We have
2q = p + r
[
a3 + b3 = (a + b)3 – 3ab (a + b)
p, q, r are in A.P.]
= (a + b) [(a + b)2 - 3ab]
p = – 9r, q = – 4r
(
)
35. Note that we must have 3x2 + x + 5 ≥ 0 and x – 3
≥ 0 or x ≥ 3. Squaring both sides of (1), we get
16
1
52
- 4Ê - ˆ =
Ë
¯
81
9
81
2
fi |a - b | =
13
9
32. As roots of 2x2 + 3x
numbers, the given quadratic equations have both
roots in common. [ See Solution Question No. 26.]
=
a b c
fia:b:c=2:3:4
= =
2 3 4
33. Case 1: When x < 3/2
Thus,
3x2 + x + 5 = x2 – 6x + 9
fi
2x2 + 7x – 4 = 0
fi
(2x – 1)(x + 4) = 0
fi
x = 1/2, – 4
x ≥ 3. Thus,
(1) has no solution.
36. a + b = 6, ab = - 2.
a10 – 2a8 = a10 + b10 + ab (a8 + b8)
In this case the equation becomes
= (a + b) (a9 + b9) = 6a9
x2 – (2x – 3) – 4 = 0
fi
fi
(x – 1)2 = 2
fi
x=1±
fi
x–1=± 2
2
As x < 3/2, we take x = 1 –
}
= 280 2
2
4r
Ê -q ˆ
=Á ˜ Ë p¯
p
x2 – 2x + 1 = 2
{
= 4 2 ( 2) ÈÎ66 - 2(24 ) - 1 ˘˚
|a - b|2 = (a + b)2 - 4ab
fi
k > 0]
[
a10 - 2a8
=3
2a9
2x3 – 9x2 – kx – 13 = 0
jugate pair, so let the roots of (1) be 2 + 3i, 2 – 3i
and a, where a ΠR. Now,
2
Case 2: When x ≥ 3/2
(2 + 3i) + (2 – 3i) + a = 9/2 fi
a = 1/2
In this case the equation becomes
\ real root of the equation is 1/2.
x2 + (2x – 3) – 4 = 0
fi
i) (2 – 3i)a = 13/2 fi
(x + 1)2 = 8 fi x + 1 ± 2 2
38. Write
As x ≥ 3/2, x = –1 + 2 2
(
) (
)
sum of roots = 1 - 2 + -1 + 2 2 = 2
34. a + b = 4 2 k, ab = 2e4lnk – 1 = 2k4 –1
Now,
66 = a2 + b2 = (a + b)2 – 2ab = 32k2 – 2(2k4 – 1)
fi 33 = 16k2 – 2k4 + 1
fi 2k4 – 16k2 + 32 = 0
fi k4 – 8k2 + 16 = 0
a =
1/2]
x4 + x2 + 1 = x4 + 2x2 + 1 – x2
= (x2 + 1)2 – x2
= (x2 + x + 1) (x2 – x + 1)
The given equation, now can be written as
(x2 + x + 1) [(a – 1)(x2 – x + 1) + (a + 1) (x2 + x
+ 1)] = 0
fi (x2 + x + 1) [2ax2 + 2x + 2a] = 0
Quadratic Equations 3.57
3 4 5
42. x + y + z = 12, x y z = (0.1)(600)3
As x is real, we get
We have
ax2 + x + a = 0.
x3y4z5 = (0.1)(600)(600)2
This equation will have real and distinct roots if
2
= (60)(600)2
2
a π 0 and 1 – 4a > 0 fi a π 0, a < 1/4
= 3 32 85 5
fi a Œ ( – 1/2, 0) » (0, 1/2)
2
\
2
39. If x – 5x + 5 π 1, then x + 4x – 60 = 0
x = 3, y = 4, z = 5
Thus, x3 + y3 + z3 = 33 + 43 + 53
= 27 + 64 + 125 = 216
fi (x + 10)(x – 6) = 0 fi x = –10, 6
When x = –10 and 6, x2 – 5x + 5 π 1
If x2 – 5x + 5 = 1, ( x 2 - 5 x + 5) x
2
+ 4 x - 60
=1
Previous Years' B-Architecture Entrance
Examination Questions
2
In this case, x – 5x + 4 = 0
fi (x – 1) (x – 4) = 0 fi x = 1, 4
If x2 – 5x + 5 = –1, then x2 + 4x – 60 must be
even integer.
1.
tan 30º + tan 15º = – 2p
tan 30º tan 15º = q
2
But x – 5x + 6 = 0 fi x = 2, 3
For x = 2, x2 + 4x – 60 = – 48 is even.
1 = tan 45º =
tan 30∞ + tan 15∞
1 - tan 30∞ tan 15∞
-2 p
1- q
1 – q = – 2p
For x = 3, x2 + 4x – 60 = – 39 is odd.
fi
Thus, sum of desired values is –10 + 6 + 1 + 4 +
2 = 3.
1=
fi
q = 1 + 2p.
40. Let a be common roots of
fi
2. Let roots of the equation be 1– k and 1 + k, where
k > 0.
x2 + bx – 1 = 0 and x2 + x + b = 0.
Then a2 + ba – 1 = 0 and a2 + a + b = 0
fi (a2 + ba –1) – (a2 + a + b) = 0
fi (b – 1)a – (b + 1) = 0
b +1
fia=
if b π 1
b -1
As a π –1, b π 0
Then
2 = ( 1 – k) + (1 + k) =
fi
1=
2a
a+2
a
a+2
a.
3. |x|2 – 4|x| – 2 = 0
2
Ê b + 1ˆ
Ê b + 1ˆ
Also, Á
–1=0
+ bÁ
˜
Ë b - 1¯
Ë b - 1˜¯
fi (|x| – 2)2 = 6
fi (b + 1)2 + b(b + 1)(b – 1) – (b – 1)2 = 0
fi |x| = 2 ± 6
fi 4b + b(b + 1)(b – 1) = 0
As | x | > 0, | x | = 2 +
2
fi x = ± (2 +
3
2x + 1 - 2x - 1 = 1
Also, (2x + 1) – (2x – 1) = 2
2x + 1 + 2x - 1 = 2
(1)
(2)
3
,
2
4 x2 - 1 =
6)
(3)
AIEEE/JEE Questions.
5. ax2 – bx(x – 1) + c(x – 1)2 = 0
fi
From (1) and (3), we get
\
6
\ there are two values of x.
41. We are given
2x + 1 =
|x| – 2 = ± 6
2
As b π 0, 4 + b – 1 = 0 fi b = 3
fi |b| =
fi
2x - 1 =
3
4
1
2
fi
fi
x ˆ2
x ˆ
Ê
Ê
+ bÁ aÁ+c=0
˜
Ë x - 1˜¯
Ë x - 1¯
x
= a,b
x -1
a
b
x=
,
a +1 b +1
Complete Mathematics—JEE Main
3.58
x4– ax2 + b = (x2 – 3x + 2) p (x)
6.
9. The given equation can be written as
= (x – 1)(x – 2) p(x)
r(2x + p + q) = (x + p) (x + q)
where p(x
\1–a+b=0
fi
and
16 – 4a + b = 0
Let roots of this equation be a , -a.
a = 5, b = 4
Then
Thus, equation whose roots are a and b is
0 = a + (-a) = - (p + q – 2r) fi 2r = p + q
2
x – 9x + 20 = 0
and
bˆ
Ê
7. f ¢(x) = 2ax + b = a Ë x + ¯
2a
As a > 0,
(x – 3k)2 = 2(k – 1)
fi
f ¢(x) > 0 for x > – b/2a
Thus, f
- b/2a)
(– b/2a, •).
a(–a) = pq – r(p + q)
1
1 2
2
2
= pq - ( p + q) = - ( p + q )
2
2
10. Write the equation as
f ¢(x) < 0 for x < – b/2a
and
x2 + (p + q – 2r) + pq – r(p + q) = 0
or
•,
x = 3k ± 2 k - 1
Both the roots will be greater than 3 if smaller root
is greater than 3, that is, if
3k - 2 k - 1 > 3
-b
4ac - b 2
Also, Min f(x) = f Ê ˆ =
>0
Ë 2a ¯
4a
2
if b < 4ac
fi
3(k - 1) > 2 k - 1
fi
k > 1 and 9(k – 1) > 2 fi
As Min f(x) > 0, f(x) > 0 " x ΠR
fi
11 – 9k < 0.
\ Both the statements are true and statement- 2 is
k > 11/9
11. (1 + 3m)2 – (1 + m2)(1 + 8m) < 0
fi 1 + 6m + 9m2 – (1 + m2 + 8m + 8m3) < 0
8. As the roots are of opposite signs, the product of
roots must be negative, that is,
2
a - 3a + 2
< 0 fi (a - 1)(a - 2) < 0
3
fi a Π(1, 2)
fi –2m + 8m2 – 8m3 < 0
fi 2m(4m2 – 4m + 1) > 0
fi m(2m – 1)2 > 0
which is true for each m ΠN.
CHAPTER FOUR
Determinants
EVALUATION OF DETERMINANTS
MINORS AND COFACTORS
A determinant of order two is written as
Minor
a11
a21
a12
a22
(ai j ΠC " i, j)
and is equal to a11 a22 – a12 a21.
A determinant of order three is written as
a11
a21
a31
a12
a22
a32
a13
a23
a33
a22
a32
a23
a
- a12 21
a33
a31
1
Illustration
(ai j ΠC " i, j)
and is equal to
a11
Let A = (aij)n × n be a square matrix of order n. Then the minor
Mij of the element aij of the matrix A is the determinant of
the square sub-matrix of order (n – 1) obtained by deleting
ith row and jth column of matrix A.
a23
a
+ a13 21
a33
a31
a22
a32
= a11 (a22 a33 – a23 a32) – a12 (a21 a33 – a23 a31)
+ a13(a21 a32 – a22 a31)
= a11 a22 a33 + a12 a23 a31 + a13 a32 a21
– a13 a31 a22 – a32 a23 a11 – a12 a21 a33
A determinant of order 3 can also be evaluated by using
the following diagram, due to Sarrus:
Minor of element a23 in the determinant
a11 a12 a13
a21 a22 a23
a31 a32 a33
is
a
a
M23 = 11 12
a31 a32
Cofactor
Let A = (aij)n ¥ n be a square matrix of order n. Then the
cofactor of the element aij of the matrix A is denoted by Cij
and is equal to (- 1)i + j Mij where Mij is the minor of the
element aij of the matrix A.
Note that
a11 a12 a13
a21 a22 a23 = a11 M11 – a12 M12 + a13 M13
= a11 C11 + a12C12 + a13C13
a
a
a
31
The product of the three terms on each of the three single
arrows are prefixed by a positive sign and the product of the
three terms on each of the three double arrows are prefixed
by a negative sign.
32
33
a11 a12 a13
D = a21 a22 a23
a31 a32 a33
D = ai1 Ci1 + ai2 Ci2 + ai3 Ci3
= a1j C1j + a2j C2j + a3j C3j
If
then
i = 1, 2, 3
j = 1, 2, 3
Remark
Note
This method does not work for higher order determinants.
and
ai1 Cj1 + ai2 Cj2 + ai3 Cj3 = 0
a1i C1j + a2i C2j + a3i C3j = 0
iπj
iπj
4.2
Complete Mathematics—JEE Main
The above results remain true for determinants of every
order.
To evaluate
1+ a
b
c
D= a
1+ b
c
a
b
1+ c
D = D1 + aD2 where
PROPERTIES OF DETERMINANTS
write
1.
Reflection Property
The determinant remains unaltered if its rows are changed
into columns and the columns into rows.
In other words, if A is a square matrix, then |A| = |A¢| where
A¢ is transpose of A.
2.
1
b
c
D1 = 0 1 + b
c
0
b
1+ c
= (1 + b) (1 + c) – bc = 1 + b + c
and
All-zero Property
1
b
c
c
D2 = 1 1 + b
1
b
1+ c
If all the elements of a row (column) are zero, then the determinant is zero.
3.
Proportionality [Repetition] Property
If the elements of a row (column) are proportional [identical] to the element of the some other row (column), then the
determinant is zero.
4.
Switching Property
The interchange of any two rows (columns) of the determinant changes its sign.
5.
Scalar Multiple Property
If all the elements of a row (column) of a determinant are
multiplied by a non-zero constant, then the determinant gets
multiplied by the same constant.
6.
Property of Invariance
a1
a2
a3
b1
b2
b3
c1
a1 + a b1 + b c1
c2 = a2 + a b2 + b c2
c3
a3 + a b3 + b c3
b1
b2
b3
c1
c2
c3
d1
a1
d2 = a2
d3
a3
c1
c2
c3
d1
b1
d2 + b2
d3
b3
8.
Factor Property
If a determinant D becomes zero when we put x = a, then
(x – a) is a factor of D.
9.
Triangle Property
If all the elements of a determinant above or below the main
diagonal consists of zeros, then the determinant is equal to
the product of diagonal elements. That is,
a1
0
0
Sum Property
a1 + b1
a2 + b2
a3 + b3
Using C2 Æ C2 – bC1 and C3 Æ C3 – cC1,
we get
1 0 0
D2 = 1 1 0 = 1
1 0 1
\
D =1 + a + b + c
c1
c2
c3
That is, a determinant remains unaltered under an operation of the form Ci Æ Ci + a Cj + b Ck, where j, k π i, or an
operation of the form Ri Æ Ri + a Rj + b Rk, where j, k π i.
7.
2
Illustration
Remark
c1
c2
c3
d1
d2
d3
Remark
It is one of the most under used property. But evaluation of
some of the determinants beame very easy when we use it.
10.
a2
b2
0
a3
a1
b3 = a2
c3
a3
0
b2
b3
0
0 = a1 b2 c3
c3
Product of Two Determinants
a1
a2
a3
b1
b2
b3
c1
c2
c3
a1
a2
a3
a1a1 + b1b1 + c1g 1
= a2a1 + b2 b1 + c2g 1
a3a1 + b3 b1 + c3g 1
b1
b2
b3
g1
g2
g3
a1a 2 + b1b2 + c1g 2
a2a 2 + b2 b2 + c2g 2
a3a 2 + b3 b2 + c3g 2
a1a 3 + b1b3 + c1g 3
a2a 3 + b2 b3 + c2g 3
a3a 3 + b3 b3 + c3g 3
Determinants 4.3
Here we have multiplied rows by rows. We can also multiply rows by columns, or columns by rows, or columns by
columns.
11.
where a21, a22, a23, a31, a32 and a33 are constants, then
a11
¢ ( x ) a12
¢ ( x ) a13
¢ ( x)
D ¢(x) = a21
and
a22
a23
a31
a32
a33
Conjugate of a Determinant
If ai, bi, ci ΠC (i = 1, 2, 3), and
a1
Z = a2
a3
12.
b1
b2
b3
a1
c1
c2 then Z = a2
a3
c3
b1
b2
b3
Ú a11 ( x)d x Ú a12 ( x)d x Ú a13 ( x)d x
c1
c2
c3
Differentiation of a Determinant
Ú D( x)dx =
a1 ( x ) a2 ( x )
a3 ( x ) a4 ( x )
D(x) =
then
a ¢ ( x ) a2 ( x )
a ( x ) a2¢ ( x )
D¢(x) = 1
+ 1
a3¢ ( x ) a4 ( x )
a3 ( x ) a4¢ ( x )
If we write D(x) = [C1, C2], where Ci denotes ith column,
then
D¢(x) = [C 1¢ , C2] + [C1, C 2¢ ] where C 1¢ denotes the column
which contains the derivative of all the functions in the ith
column Ci. Similarly, if
È R¢ ˘ È R ˘
ÈR ˘
D(x) = Í 1 ˙ then D¢(x) = Í 1 ˙ + Í 1 ˙
R
Î R2 ˚ Î R2¢ ˚
Î 2˚
a11 ( x ) a12 ( x ) a13 ( x )
D(x) = a21 ( x ) a22 ( x ) a23 ( x )
a31 ( x ) a32 ( x ) a33 ( x )
D¢(x)
Next, if
then
=
a11
a11 ( x ) a12
¢ ( x ) a12 ( x ) a13 ( x )
¢ ( x ) a13 ( x )
a21
¢ ( x ) a22 ( x ) a23 ( x ) + a21 ( x ) a22
¢ ( x ) a23 ( x )
a31
a31 ( x ) a32
¢ ( x ) a32 ( x ) a33 ( x )
¢ ( x ) a33 ( x )
a11 ( x ) a12 ( x ) a13
¢ ( x)
+ a21 ( x ) a22 ( x ) a23
¢ ( x)
a31 ( x ) a32 ( x ) a33
¢ ( x)
= [C 1¢ , C2, C3] + [C1, C 2¢ , C3] + [C1, C2, C ¢3]
Similarly, if
Corollary
If
È R1 ˘
È R1¢ ˘ È R1 ˘ È R1 ˘
Í
˙
D(x) = R2 then D¢(x) = Í R2 ˙ + Í R2¢ ˙ + Í R2 ˙
Í ˙
Í ˙ Í ˙ Í ˙
ÍÎ R3 ˙˚
ÍÎ R3 ˙˚ ÍÎ R3 ˙˚ ÍÎ R3¢ ˙˚
(Differentiation and Integration of Determinant)
a11 ( x ) a12 ( x ) a13 ( x )
a22
a23
D(x) = a21
a31
a32
a33
a22
a32
a23
a33
In general, for any positive integer m
m
m
m
a11
( x ) a12
( x ) a13
( x)
m
D (x) = a21
a22
a23
a31
a32
a33
If each ai(x) is differentiable function
and
a21
a31
13.
Determinant of Cofactor Matrix
a11
If D = a21
a31
C11 C12
a13
a23 then D1 = C21 C22
C31 C32
a33
a12
a22
a32
C13
C23 = D2
C33
where Ci j denotes the co-factor of the element ai j in D.
SOME TIPS FOR QUICK EVALUATION
OF DETERMINANTS
1. If D is a skew symmetric determinant of odd order,
then D = 0
Illustration
3
0
a b
D = -a 0 c
-b -c 0
Using the reflection property, write
0 - a -b
D = a 0 -c
b c
0
Taking –1 common from R1, R2 and R3 we get
0
a b
3
D = (–1) - a 0 c = –D
-b -c 0
fi
2D = 0 fi D = 0
2. If a1, a2, a3 are in A.P.; b1, b2, b3 are in A.P. and c1, c2,
c3 are also in A.P. Then
a1 a2 a3
D = b1 b2 b3 = 0
c1 c2 c3
Use C1 Æ C1 + C3 – 2C2
Let
Complete Mathematics—JEE Main
4.4
3. If a1, a2, a3 are in G.P.; b1, b2, b3 are also in G.P., with
the same common ratio, then
a1 a2 a3
D = b1 b2 b3 = 0
c1 c2 c3
[c1, c2, c3 can be any three complex numbers.]
Some Frequently used Determinants
1.
1
a
1
b
a2
b2
1
1 1 1
c = a b c
c 2 bc ca ab
a1
a
D= 2
a3
2.
1
b
3
3
a
b
1
c = (a – b) (b – c) (c – a) (a + b + c)
c
c1
c2
c3
If D π 0, then the only solution of the above system of
equations is x = 0, y = 0 and z = 0.
Corollary If at least one of x, y, z is non-zero and x, y and
z are connected by the three given equations, then the elimination of x, y and z leads to the relation
a1
a2
a3
= (a – b) (b – c) (c – a)
1
a
b1
b2
b3
b1
b2
b3
c1
c2 = 0
c3
CRAMER’S RULE
3
1
1
1
3. a 2
b2
c 2 = (a – b) (b – c) (c – a) (bc + ca + ab)
a3
b3
c3
If
a b c
4. b c a = 3abc – a3 – b3 – c3
c a b
= (a + b + c) (bc + ca + ab – a2 – b2 – c2)
1
= - (a + b + c)[(b - c)2 + (c - a)2 + (a - b)2 ]
2
LINEAR EQUATIONS
The system of linear homogeneous equations
a1 x + b1 y + c1 z = 0
a2 x + b2 y + c2 z = 0
a3 x + b3 y + c3 z = 0
has a non-trivial solution (i.e., at least one of the x, y, z is
different from zero) if and only if D = 0, where
a1
D = a2
a3
b1
b2
b3
c1
c2 π 0
c3
then the solution of the system of linear equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3
is given by x =
D
D1
D
,y= 2 ,z= 3
D
D
D
where
d1
D1 = d2
d3
b1
b2
b3
c1
a1
c2 , D 2 = a2
c3
a3
a1
D 3 = a2
a3
b1
b2
b3
d1
d2
d3
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Example 1: Let
1
x
D(x, y) = 1 x + y
1
x
y
y
x+y
Then D(–3, 2) equals
(a) 13
(c) 12
Ans. (b)
(b) – 6
(d) – 5
d1
d2
d3
c1
c2 and
c3
Determinants 4.5
Solution: Using R2 Æ R2 – R1, R3 Æ R3 – R1 we get
1 x y
D(x, y) = 0 y 0 = xy
0 0 x
\
D(–3, 2) = – 6
Example 2: Suppose a, b, c are distinct real numbers and
a a2
D= b b
2
c c2
Then a + b + c equals
(a) –1
(c) 0
Ans. (b)
b+c
c+a =0
a+b
(b) 2
(d) –5
Solution: Using C3 Æ C3 + C1 and taking (a + b + c)
common from C3 we get
(1)
D = (a + b + c) D1
where
D1 = b b 2 1
c2
1
Using R1 Æ R1 – R2 and R2 Æ R2 – R3, we get
a - b a 2 - b2
D1 = b - c
c
Solution: Interchanging the rows and columns, we get
0
a-b a-c
0
b-c
D = b-a
c-a c-b
0
Taking –1 common from each of R1, R2, R3 we get
0
b-a c-a
3
0
c-b
D = (–1) a - b
a-c b-c
0
fi
D = –D fi 2D = 0 or D = 0
Alternative Solution
D is a skew symmetric determinant of odd order,
therefore D = 0
a a2 1
c
Example 4: Let
0
b-a c-a
D = a-b
0
c - b , then
a-c b-c
0
D equals:
(a) 0
(b) abc
(c) a2 + b2 + c2
(d) bc + ca + ab
Ans. (a)
2
b -c
0
2
0
c2
1
1 a+b 0
= (a – b) (b – c) 1 b + c 0
c
= (a – b) (b – c)
c2
1
1 a+b
1 b+c
[Expand along C3]
= (a – b) (b – c) (c – a)
(2)
From (1) and (2)
D = (a + b + c) (a – b) (b – c) (c – a)
As
D = 0 and a, b, c are distinct,
we get
a+b+c=0
Example 3: Suppose A = (aij)3 × 3, where aij Œ R. If
det (adj A) = 25, then |det (A)| equals:
(a) 5
(b) 12.5
(d) 52/3
(c) 5 5
Ans. (a)
Solution: Using det (adj A) = (det (A))2, we get
(det (A))2 = 25 fi |det (A)| = 5
Example 5: Let A = (aij)3 × 3, where aij Œ C the set of
complex numbers. If det (A) = 2 – 3i, then det (A–1) equals:
1
1
(a)
(2 – 3i)
(b)
(2 + 3i)
13
13
(c) 2 – 3i
(d) 2 + 3i
Ans. (b)
1
1
=
Solution: det (A–1) =
det ( A) 2 - 3i
2 + 3i
1
=
= (2 + 3i )
2
2
13
2 +3
Example 6: In a triangle ABC, if
1 a b
D= 1 c a =0
1 b c
2
then sin A + sin B + sin2C is:
3 3
2
9
(c)
4
Ans. (a)
(a)
2
(b)
5
4
(d) 2
Solution: Evaluating along C1, we get
D = (c2 – ab) + (b2 – ac) + (a2 – bc) = 0
fi
(a – b)2 + (b – c)2 + (c – a)2 = 0
fi
a = b = c fi A = B = C = p /3
\
sin2A + sin2B + sin2C =
3 3
.
2
Complete Mathematics—JEE Main
4.6
Example 7: If
a - b a 2 - b2
2
x +x
x +1
2
2 x + 3x - 1
x-2
3 x - 3 = Ax + B,
3x
2
x + 2x + 3
D1 = b - c
c
2x - 1 2x - 1
then A is equal to:
(a) 12
(c) 24
Ans. (c)
Ax + B =
x-2
0
x2 + 2 x + 3 2 x - 1 2 x - 1
x +1 x - 2
2x - 1 2x - 1
3 x-2
0 2x - 1
= 4(3) (2x – 1) = 24x – 12
A = 24
=4
Thus,
Let
p + a3
D = b b2
p + b3
c2
p + c3
c
Now,
1
D1 = b b 2 1
c
c2
a a
and
2
b3
c2
c3
c
write
a a2 1
D2 = abc 1 b b2 = abc(-1)2 b b2 1
1 c
c2
a
3
D2 = b b 2
1
1 a a2
0 -51 -71
– c = - P(0) = - 51 0 -73
71 73
0
=0
[ Skew symmetric determinant of odd order]
Example 10: Let
Solution: Write D = pD1 + D2
where
a a
TIP
Suppose P(x) = x3 + ax2 + bx + c, then product of zeros
of P(x) is – c = –P(0).
If D = 0, then which one of the following is not true
(a) a = –1, b = 1
(b) b = 1, c = p
(c) a = 0, c = p
(d) abc + p = 0
Ans. (c)
2
1
Solution:
Example 8: Suppose a, b, c are three integers such a < b
< c and p is a prime number.
a a2
0
Example 9: Suppose
x -51 -71
P(x)= 51 x -73
71 73
x
Product of zeros of P(x) is
(a) 0
(b) 195
(c) –195
(d) –264333
Ans. (a)
Expanding along R2, we get
Ax + B = (-1)(- 4)
c2
0
1 a+b
1 b+c
= (a – b) (b – c) (c – a)
As a < b < c, D1 π 0. Therefore,
D = 0 fi p + abc = 0
fi
p = –abc
As p is prime and a, b, c are integers such that a < b < c, we
must have – a = 1, b = 1, c = p.
fi
a = –1, b = 1, c = p.
Solution: Applying R2 Æ R2 – R1 – R3, we get
x +1
0
b -c
2
= (a - b)(b - c)
(b) 18
(d) 30
x2 + x
-4
2
c
= abc D1
Thus,
D = (p + abc) D1.
Using R2 Æ R2 – R3, R1 Æ R1 – R2, we get
c2
1
x -3 + 4i 3 - 4i
-7i
5 + 6i
P(x) = x
- x 7 - 2i -7 - 2i
The number of values of x for which P(x) = 0 is
(a) 0
(b) 1
(c) 2
(d) 3
Ans. (b)
Solution: Using R2 Æ R2 – R1, R3 Æ R3 + R1,
we get
x -3 + 4i 3 - 4i
P(x) = 0 3 - 11i 2 + 10i
0 4 + 2i -4 - 6i
As P(x) is a linear polynomial, P(x) = 0 for exactly one
value of x.
Determinants 4.7
Example 11: Let
Thus,
sin q
1
- sin q
1
D (q) = - sin q
-1
Solution of D(q) = 3 is
p 3p
(a)
,
2 2
(c)
{ }
{ }
1
sin q , 0 £ q £ 2p
1
{
{
(b)
p 3p
,
4 4
(d)
p 3p 5p 7p
, , ,
4 4 4 4
p p 3p
, , ,p
4 2 4
}
= (1 + a2 + a4) x2 – x3
As x π 0, D(x) = 0
1
sin q
1
= 2 ( 1 + sin2q )
D(q) = 3 fi 2 sin2q = 1
p 3p 5p 7p
1
sin q = ±
fi q= , , ,
4 4 4 4
2
\
fi
a
a2
a
a2 - x
a3
D(x) =
a2
a3
a4 - x
Number of values of x for which D(x) = 0 is
(a) 0
(b) 1
(c) 2
(d) 3
Ans. (b)
Solution: Using the sum property, write
D(x) = D1 –xD2
where
a
a2
a2 - x
a3
a2
a3
a4 - x
1
a
a2
1
D1 = a
and
2
D2 = 0 a - x
0
a
a
3
2
3
4
a -x
4
= (a – x) (a – x) – a6
= – (a2 + a4)x + x2
In D1, use C2 Æ C2 – aC1, C3 Æ C3 – a2C1 to obtain
1
D1 = a
0
-x
0
0 = x2
a2
0
-x
a2
b2
c2
( a + l )2
( b + l )2
( a - l )2
( b - l )2
a2
( c + l )2 = k l a
1
( c - l )2
l π 0, then k is equal to:
(a) 4 labc
(c) 4 l2
Ans. (c)
b2
b
1
c2
c
1
(b) – 4 labc
(d) – 4 l2
Solution: Using R3 Æ R3 – R2 and R2 Æ R2 – R1,
we get
a2
D = 2 al + l 2
- 4 al
Example 12: Suppose a Œ R and x π 0. Let
1- x
1
x=
Example 13: If
Ans. (b)
2 sin q
1
D(q) = 0
0 - sin q
fi
.
1 + a + a2
Thus, D(x) = 0 for exactly one value of x.
}
Solution: Using C1 Æ C1 + C3, we get
D(x) = x2 + (a2 + a4) x2 – x3
b2
c2
2 bl + l 2
- 4 bl
2cl + l 2
- 4cl
Take – 4l common from R3, and applying R2 Æ R2 – 2lR3,
we get
a 2 b2 c 2
D = - 4l 3 1 1 1
a b c
a2
= l (4l ) a
1
2
\
b2
b
1
c2
c
1
k = 4l2
Example 14: Let
1
f (q) = - sin q
-1
cos q
1
sin q
-1
- cos q
1
Suppose A and B are respectively maximum and minimum
value of f (q). Then (A, B) is equal to:
(a) (2, 1)
(b) (2, 0)
(d) ÊÁ 2, 1 ˆ˜
(c) ( 2 , 1)
Ë
2¯
Ans. (b)
Solution: Using C1 Æ C1 + C3, we get
0
cos q
1
f (q) = - (sin q + cos q )
0
sin q
Evaluating along C1, we get
-1
- cos q
1
Complete Mathematics—JEE Main
4.8
f (q) = (sin q + cos q )
cos q
sin q
has a non-trivial solution, then ab + bc + ca equals:
(a) a + b + c
(b) a b c
(c) 1
(d) –1
Ans. (b)
-1
1
= (sin q + cos q)2
2
= È 2 ÊÁ 1 sin q + 1 cos q ˆ˜ ˘
ÍÎ Ë 2
¯ ˙˚
2
Êp
ˆ
= 2 sin 2 Ë + q ¯
4
As
and
Thus,
Êp
ˆ
0 £ sin 2 Ë + q ¯ £ 1, we get
4
0 £ f (q) £ 2
p
A = 2 for q =
4
p
B = 0 for q = 4
(A, B) = (2, 0)
Example 15: If a, b, c are non-zero real numbers and if
the system of equations
(a – 1)x = y + z,
(a – 1)y = z + x,
(a – 1)z = x + y,
Solution: As the given system of equations has a nontrivial solution,
a - 1 -1
-1
D = -1 b - 1 -1 = 0
-1
-1 c - 1
Write
a -1
-1
1 -1
-1
D = 0 b - 1 -1 - 1 b - 1 -1
0 -1 c - 1 1 -1 c - 1
1 0 0
= a [(b - 1)(c - 1) - 1] - 1 b 0
1 0 c
[use C2 Æ C2 + C1
C 3 Æ C 2 + C 1]
= a(bc – b – c) – bc
As D = 0, we get
ab + bc + ca = abc
LEVEL 1
Straight Objective Type Questions
Example 16: Let
x - 10
7
6
P(x) = 2
x - 10
5
x - 10
3
4
sum of zeros of P(x) is
(a) 30
(b) 28
(c) 27
(d) 25
Ans. (a)
Solution: Using R1 ´ R3, write
P(x) = –
x - 10
3
4
2
x - 10
5
7
6
x - 10
=–
coefficient of x 2
coefficient of x 3
Example 17: Let
P(x) =
x 2 - 13
4
3
x 2 - 13
6
5
= 30.
2
7
2
x - 13
If x = – 2 is a zero of P(x), then sum of the remaining five
zeros is
(a) –2
(b) 0
(c) 2
(d) 3
Ans. (c)
Solution:
TIP
TIP
Observe P(x) is of the form
P(x) = – (x –10)3 + a (x – 10) + b
where a, b are some real numbers.
\ sum of zeros of P(x)
Observe
P(x) = (x2 –13)3 + a (x2 – 13) + b.
where a, b are some real numbers.
As coefficient of x5 in P(x) is 0, sum of six zeros of P(x) is 0.
fi sum of the remaining five zeros + (–2) = 0
Determinants 4.9
fi
sum of the remaining five zeros = 2.
=
a, b are two real numbers and
Example 18: Suppose
f(n) = a n + b n. Let
3
D = 1 + f (1)
1 + f (2 )
1 + f (1) 1 + f (2)
1 + f (2) 1 + f (3)
1 + f (3) 1 + f (4)
If D = k (a – 1)2 (b – 1)2 (a – b)2, then k is equal to
(a) 1
(b) 4 ab
(c) 9
(d) a2 b2
Ans. (a)
Solution:
1+a + b
1+ a2 + b2
1+ a2 + b2
1 + a3 + b3
1+1+1
D = 1+a + b
1+ a2 + b2
1
= 1
1
a
1 a2
where
1
D1 = 1
1 a
2
1
a
b2 1 a2
1
a
1
b = D12
0
a -1
\
D = (a + b + c) (a2 + b2 + c2) = 0
As a, b, c are distinct real numbers, a2 + b2 + c2 π 0, therefore
a + b + c = 0.
fi the line a (x – 5) + b(y – 2) + c = 0 passes through (6, 3).
Example 20: Suppose a, b, c and x are real numbers.
Let
1 + a 1 + ax 1 + ax 2
D = 1 + b 1 + bx 1 + bx 2
1 + c 1 + cx 1 + cx 2
2
Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we get
1
D1 = 1
a c -b
2
2
2
D1 = b 0
a = a (a + b + c )
c -a 0
b2
1
b
b
a a+c a-b
D1 = b
b
a+b
c c-a
c
Using C2 Æ C2 – C1, C3 Æ C3 – C1, we get
1 + a3 + b3 1 + a 4 + b4
1 1
b 1
0
b -a
Then D is independent of
(a) a, b, c
(c) a, b, c, x
Ans. (c)
1 1 + ax 1 + ax 2
D1 = 1 1 + bx 1 + bx 2
1
1
= (a - 1) ( b - a )
a +1 b +a
= (a –1) (b –1) (b – a)
D = (a –1)2 (b –1)2 (a – b)2
k =1
Example 19: Suppose a, b and c are distinct real
numbers. Let
a
a+c a-b
D = b-c
b
a+b = 0
c+b c-a
c
Then the straight line a(x – 5) + b(y – 2) + c = 0 passes
through the fixed point
(a) (5, 2)
(b) (6, 2)
(c) (6, 3)
(d) (5, 3)
Ans. (c)
Solution: Applying C1 Æ aC1 + bC2 + cC3, we get
Let
a (a + b + c) a + c a - b
1
b
a+b
D = b (a + b + c)
a
c (a + b + c) c - a
c
(b) x
(d) none of these
Solution: Write D = D1 + D2, where
1 a2 -1 b2 - a2
Thus,
\
1
a (a + b + c) D1 , where
a
1 1 + cx 1 + cx 2
a ax ax 2
and
D2 = b bx bx 2 = 0
cx 2
[ C1 and C2 are propotional]
In D1, use C2 Æ C2 – C1, C3 Æ C3 – C1 to obtain
c
cx
1 ax ax 2
D1 = 1 bx bx 2 = 0
cx 2
[ C2 and C3 are proportional]
D = 0 and hence independent of a, b, c, x.
1 cx
Thus,
Example 21: Suppose a, b, c > 1 and
a- x
ax
-3 x
3x
f (x)= b
c -5 x
b
c5 x
x
3x3 , x ΠR
5x5
4.10
Complete Mathematics—JEE Main
then f is
(a)
(b)
(c)
(d)
Ans. (d)
1
b
c
D1 = 0 c - b a - c
0 a-b b-c
= – (b – c)2 – (a – b) (a – c)
= – (a2 + b2 + c2 – bc – ca – ab)
1
fi
D1 = – [(b – c)2 + (c – a)2 + (a – b)2] < 0
2
As a + b + c > 0, we get
D = (a + b + c) D1 < 0
a constant function
a polynomial of degree 5
an odd function
an even function
Solution:
ax
a- x
3x
-3 x
-3 x 3
c -5 x
-5 x 5
f (–x) = b
b
c5 x
-x
Example 24: Suppose A, B, C are angles of a triangle,
and let
e2iA e-iC e-iB
a- x
ax
x
= (-1)(-1) b -3 x
b3 x
3x3
-5 x
5x
5
c
c
5x
= f(x)
Thus, f is an even function.
Example 22: Suppose n, m are natural numbers and
1
f(x) = (1 + mx )n
(1 + nx )m
(1 + x )m
(1 + mx )mn
1
(1 + nx )mn
(1 + x )n
1
constant term of the polynomial f (x) is:
(a) 1
(b) m + n
(c) m – n
(d) 0
Ans. (d)
Solution: Constant term of polynomial f (x) is f (0), and
1 1 1
f (0) = 1 1 1 = 0
Solution: Using C1 Æ C1 + C2 + C3, we get
D = (a + b + c) D1
where
1 b c
D1 = 1 c a
1 a b
Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get
e2iB
e-iA
e-iB
e-iA
e2iC
Then value of D is
(a) – 1
(b) – 4
(c) 0
(d) 4
Ans. (b)
Solution: Taking eiA, common from R1, eiB from R2 and
iC
e from R3, we get
D = ei(A + B + C) D1
where
eiA
e-i( A + C ) e-i( A + B)
D1 = e - i ( B + C )
e-i( B + C )
eiB
e-i( A + B)
e-i( A + C )
eiC
But A + B + C = p, so that ei(A + B + C) = eip
= cos p + i sin p = –1. Also,
A + C = p – B fi e–i(A + C) = e–pi eiB = –eiB.
eiA
1 1 1
Example 23: Suppose a, b, c are sides of a scalene
triangle. Let
a b c
D= b c a
c a b
Then
(a) D £ 0
(b) D < 0
(c) D > 0
(d) D ≥ 0
Ans. (b)
D = e -iC
Thus,
-eiB
-eiC
D1 = -e- iA
eiB
-eiC
-eiA
-eiB
eiC
1
=e
-1
-1
Using C1 Æ C1 + C2, we get
0 -1
D1 = (-1) 0 1
-2 -1
Therefore, D = (–1) D1 = – 4
i ( A + B +C )
-1 -1
1 -1
-1 1
-1
-1 = (–1) (–2) (2) = 4
1
Example 25: Suppose x1, x2, x3 are real numbers such
that x1 x2 x3 π 0. Let
x1 + a1b1
a1b2
a1b3
x2 + a2 b2
a2 b3
D = a2 b1
a3 b1
a3 b2
x3 + a3 b3
Determinants 4.11
Then
D
- 1 equals:
x1 x2 x3
1 -4
Example 27: Let D = 1 - 2
(b) – 1
(c)
1 2x
D = 0 is
(a) {– 2, 3}
(c) {4, – 6}
Ans. (d)
ab
ab a b
(a) 1 1 + 2 2 + 3 3
x1
x2
x3
a1a2 a3 + b1b2 b3
x1 x2 x3
1
D= 0
Solution: Using the sum property, write
D = x1 D1 + b1 D2
where
1
a1b2
a1b3
a2 b3
D1 = 0 x2 + a2 b2
0
a3 b2
x3 + a3 b3
(b) {– 3, 4}
(d) {– 2, – 1}
-4
2
20
- 15
= x2 x3 + x2 a3 b3 + x3 a2 b2
a1b2
x2 + a2 b2
a3 b2
a1b3
a2 b3
x3 + a3 b3
Now,
0
x2
0
= 10(x + 2) (x + 1)
D = 0 fi x = – 2, – 1.
Example 28: Let D = 1 b b2 - ca , then D is equal to
1 c c 2 - ab
(a) 0
1 2
(a + b2 + c2)
(c)
2
Ans. (a)
D = x1[x2 x3 + x2 a3 b3 + x3 a2 b2] + a1 b1 x2 x3
fi
D
ab
ab a b
-1 = 1 1 + 2 2 + 3 3
x1 x2 x3
x1
x2
x3
4
3
(d) none of these
1 a a 2 - bc
D = (b – a) (c – a) 0 1 a + b + c = 0
0 1 a+b+c
[
R2 and R3 are identical]
Example 29: Suppose a, b, c > 0 and a, b, c are the
pth, qth, rth terms of a G.P. Let
1 p log a
D= 1 q
2
Example 26: Let pl + ql + rl + sl + t
l 2 + 3l
= l -1
l -3
(b) a + b + c
Solution: Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get
0
0 = a1 x2 x3
x3
Thus,
1
- 15
1 5 ( x - 2)
1 a a 2 - bc
Using C2 Æ C2 – b2 C1, C3 Æ C3 – b3 C1, we get
a1
D2 = a2
a3
= 2(x + 2)
0 2 ( x + 2) 5 ( x 2 - 4)
= (x2 + a2 b2) (x3 + a3 b3) – a3 b2 a2 b3
a1
D2 = a2
a3
5x2
Solution: Applying R2 Æ R2 – R1, R3 Æ R3 – R1 we get
(d) 0
Ans. (a)
and
20
5 . Solution set of
l -1 l - 3
- 2l l - 4
l + 4 3l
where p, q, r, s and t are constants. Then value of t is
(a) 0
(b) – 1
(c) 2
(d) 3
Ans. (a)
Solution: Putting l = 0, we obtain
0 -1 3
t= 1
0 -4 = 0
-3 4
0
as it is a skew symmetric determinant of odd order.
log b
1 r log c
then numerical value of D is
(a) – 1
(b) 2
(c) 0
(d) none of these
Ans. (c)
Solution: Let a = ARp – 1, b = AR q – 1 and c = ARr – 1
fi
log a = a + (p – 1)b, log b = a + (q – 1)b and
log c = a + (r – 1)b
where
a = log a, b = log R.
Now,
1 p a + ( p - 1) b
D = 1 q a + (q - 1) b
1 r
a + (r - 1) b
Using C3 Æ C3 – (a – b)C1 – bC3, we get D = 0.
4.12
Complete Mathematics—JEE Main
1
– 1 + 3i and
2
(
Example 30: Let w =
1
)
1
1
2
D= 1 -1-w
w
1
w 2 then D equals
2
w
(a) 3w
(c) 3w2
Ans. (b)
Solution: If n is a multiple of 3, we get each element
of D becomes 1.
\
D = 0.
If
n = 3k + 1, then
4
(b) 3w (w – 1)
(d) 3w (1 – w)
1
D= 0
w
D= w
w2
If
w 2 = 3(w2 – w) = 3w (w – 1)
Then numerical value of D is
(a) 0
(b) – 1
(c) 1
(d) none of these
Ans. (a)
1 1 1
, , are the pth, qth, rth terms of an A.P.
a b c
1
1
= A + (p – 1)D,
= A + (q – 1)D,
Let
a
b
1
r
A + (r - 1)D
w 2n
D = wn
w 2n
1
1
wn
w 2n
Then D equals
(a) 0
(c) w
Ans. (a)
(b) 1
(d) w2
1 = 0 [use C1 Æ C1 + C2 + C3]
w
1
w2
y+z
x a p
z + x and D1 = y b q
x+y
z c r,
then
(a) D = 2D1
(c) D = 4D1
Ans. (a)
(b) D = – 2D1
(d) D = – 4D1
Solution: Using R1 Æ R1 + R2 + R3, we get
2(a + b + c) 2( p + q + r ) 2( x + y + z )
D=
c+a
r+p
z+x
a+b
p+q
x+y
Taking 2 common from R1 and applying R2 Æ R2 – R1,
R3 Æ R3 – R1, we obtain
a+b+c
-b
D =2
-c
p+q+r
-q
-r
x+y+z
-y
-z
Applying R1 Æ R1 + R2 + R3, we get
a p x
a
D = 2(– 1) (– 1) b q y = – 2 b
c r z
c
x p
y q
z r
x a p
= 2 y b q = 2D1
z c r
Example 32: Let w π 1 be complex cube root of unity
and n be a natural number and
wn
w
b+c q+r
D = c+a r+ p
a+b p+q
Using R3 Æ R3 – (A – D)R1 – DR2, we get
D = 0.
1
w
Example 33: Let
Now
1
q
A + (q - 1)D
w
1
2
D= w
1
and = A + (r – 1)D.
c
1
D = abc
p
A + ( p - 1)D
1 = 0 [use C1 Æ C1 + C2 + C3]
w2
1
w
Example 31: Let a, b, and c be respectively the pth, qth
and rth terms of a harmonic progression and
1 1 1
D= p q r
bc ca ab
2
w2
n = 3k + 2, then
1
0 w2
Solution:
w
w
Solution: Using w3 = 1 and 1 + w + w2 = 0, and applying C1 Æ C1 + C2 + C3, we get
3
1
x+y
x
x
Example 34: If x = – 2, and D = 5 x + 4 y 4 x 2 x
10 x + 8 y 8 x 3 x
then numerical value of D is
(a) 8
(c) 4
Ans. (b)
(b) – 8
(d) – 4
Determinants 4.13
Solution: Taking x common from C2 and C3 we obtain
1 1 1
x+y
1 1
3 5
4 2
D = x 5x + 4 y 4 2 = x
10
8
3
10 x + 8 y 8 3
[using C1 Æ C1 – yC2]
Next using C1 Æ C1 – C2, C2 Æ C2 – C3, we obtain
Solution: Using change of base formula
log b
logab =
, a, b > 0, a, b π 1,
log a
2
we can write
log x log y log z
1
log x log y log z = 0.
D=
log x log y log z
log x log y log z
0 0 1
1 2
D = x 1 2 2 = x3
= x3
2 5
2 5 3
Example 37: Let w π 1 be a cube root of unity and
3
As
Example 35: If a = w π 1, is a cube root of unity, b =
– 785, c = 2008 i, and
a
a+b
a+b+c
D = 2 a 3a + 2b 4 a + 3b + 2c
3a 6a + 3b 10a + 6b + 3c
then D equals
(a) – i
(b) i
(c) 1
(d) 1 – wi
Ans. (c)
1 a+b
a+b+c
D = a 2 3a + 2b 4a + 3b + 2c
3 6a + 3b 10a + 6b + 3c
and apply C2 Æ C2 – bC1, C3 Æ C3 – cC1 to obtain
1 a
a+b
D = a 2 3a 4a + 3b
3 6a 10 a + 6c
w - w2 - 1
2
then D equals
(a) – w
(c) 0
Ans. (c)
2w
2
2w
2
2
w -1-w
(b) 3w (1 – w)
(d) 1 – w2
Solution: Using R1 Æ R1 + R2 + R3 and 1 + w + w2 =
0, we obtain
D = 0.
Example 38: Suppose x = –
y = cos
1
1 + 7i and
3
(
)
p
p
+ isin .
4
4
1
x
D= 1 x + y
1
x
Let
x
y ,
x+y
then D equals
Take a common from C2 and apply C3 Æ C3 – bC2 to obtain
1 1 1
1 1 a
3
D = a 2 3 4a = a 2 3 4
3 6 10
3 6 10a
2
Apply C3 Æ C3 – C2, C2 Æ C2 – C1 to obtain
1 0 0
3
D = a 2 1 1 = a3 = w3 = 1.
3 3 4
Example 36: Let x, y, z be positive and x, y, z π 1. Let
1
log x y log x z
1
log y z ,
D = log y x
then numerical value of D is
(a) – 1
(c) 1
Ans. (b)
2w
2w
Solution: Write
log z y
2
D=
x = – 2, D = – 8.
log z x
1 - w - w2
1
(b) 0
(d) none of these
(a) - 7
(c) i
Ans. (c)
(b) 7
(d) – 1.
Solution: Using R2 Æ R2 – R1 and R3 Æ R3 – R1, we get
1 x
D= 0 y
0 0
x
p
p
y - x = y2 = cos + i sin = 0 + i(1) = i
2
2
y
[Using De Moivre’s Theorem]
Example 39: Let x = cos
p
p
+ i sin and
3
3
1
x
x2
D = x2
1
x
2
1 x
1
then numerical value of D is
(a) 0
(b) – 1
(c) 8
(d) – 4
Ans. (c)
4.14
Complete Mathematics—JEE Main
Example 42: Let f : N Æ N be defined by
Solution: Use C3 Æ C3 – xC2 and C2 Æ C2 – xC1.
1
D = x2
0
0
1 – x3
0
f (x) = ( x + 1)2 + x - ÈÎ ( x + 1)2 + ( x + 1) ˘˚
([ ] denotes the greatest integer function). Suppose a, b, c
are three distinct natural numbers. Let
2
x – x 1 – x3
1
= (1 – x3)2 = [1 – cosp – isinp]2 = 8
[De Moivre’s Theorem]
Example 40: Let f(x) = [2 - x [2 x ]], x ΠR. ([ ] denotes
the greatest integer function.) Let x1 = 0, x2 = log23 and x3 =
2 . Suppose 0 < x4 < 1.
2
f ( x1 )
D = f ( x4 )
f ( x2 )
f ( x2 )
f ( x4 )
f ( x3 )
2
f ( x3 )
f ( x2 ) , then D is equal to:
f ( x1 )
(a) –1
(c) 1
Ans. (b)
(b) 0
(d) 2 log23
2
2
[2 ]
2x
range f (x) is {0, 1}
Also
x2
x2
2
£ 1. Therefore,
x2
[2 ] = 2 ¤ 2 is a positive integer.
Thus, f (x1) = 1, f (x2) = 1, f (x3) = 1 and f (x4) = 0.
Therefore,
1 1 1
D= 0 0 1 = 0
1 1 1
w
w
w2
1
x +w
D=
1 + x = 0, then value of x is
w2
(b) 1
(d) none of these
Solution: Applying C1 Æ C1 + C2 + C3, we get
x + w2 + w +1
2
D = w +w +1+ x
1+ x + w + w2
x
w
= x
w2
x
x +w
w
w
2
x +w
1
1+ x
w2
1
1 + x using 1 + w + w2 = 0
w2
D is clearly equal to 0 for x = 0.
D = f (b ) b
b
f (c ) c
2
c
(b) a + b + c
(d) 0
Solution: If n ΠN, then
n2 < n2 + n < (n + 1)2
n<
n2 + n < n + 1
fi
[
fi
ÈÎ ( x + 1)2 + ( x + 1) ˘˚ = x + 1 " x Œ N.
\
n2 + n ] = n
f (x) = (x + 1)2 + x Р(x + 1)2 = x " x ΠN.
Thus,
a a2
D= b b
c
a
2
b =0
2
c
[
1
(a) 0
(c) – 1
Ans. (a)
a
2
c
Example 41: If w π 1 is a cube root of unity and
x + w2
f (a) a 2
The D is equal to:
(a) – (a + b + c)
(c) –1
Ans. (d)
fi
x2
Solution: 0 £ [2x ] £ 2x fi 0 £
2
C1 and C3 are identical]
x 3 7
Example 43: If x = – 9 is a root of 2 x 2 = 0, then
7 6 x
the other two roots are
(a) 3, 7
(c) 3, 6
Ans. (b)
(b) 2, 7
(d) 2, 6
Solution: Applying R1 Æ R1 + R2 + R3, we get
x+9 x+9 x+9
x
2
D= 2
7
6
x
Taking x + 9 common from R1 and applying C2 Æ C2 – C1,
C3 Æ C3 – C1 we get
1
0
0
D = (x + 9) 2 x - 2
0
7 -1 x - 7
= (x + 9) (x – 2) (x – 7)
Thus, D = 0 fi x = – 9, 2, 7
\ the other two roots are 2 and 7.
Determinants 4.15
Solution: Multiplying R3 of D1 by xyz, we get
x b b
x b
Example 44: If D1 = a x b and D2 =
, then
a x
a a x
(a) D1 = 3 (D2)2
(c)
d
(D1) = 3D22
dx
ax
1
x2
D1 =
xyz
xyz
d
(b)
(D1) = 3D2
dx
(d) D1 = 3D3/2
2
Ans. (b)
by
a b
xyz
x y
z =
xyz
xyz
yz z x
c
z = D2
xy
2
y
xyz
[taking x, y, z common from C1, C2, C3 respectively]
D1 – D2 = 0.
fi
Example 47: If
Solution: We have
1 b b
x 0 b
x b 0
d D1
= 0 x b + a 1 b + a x 0
dx
0 a x
a 0 x
a b 1
x b
x b
x b
+
+
a x
a x
a x
= 3D2.
=
Example 45: If x e R and n e I, then the determinant
equals
cz
2
sin ( np )
sin x - cos x
log tan x
D = cos x - sin x cos [(2n + 1) p / 2 ] log cot x
log cot x
log tan x
tan ( np )
(a) 0
(c) tan (p/4 – x)
Ans. (a)
(b) log tan x – log cot x
(d) none of these
Solution: We can write D as
0
sin x - cos x log tan x
0
D = - (sin x - cos x )
- log tan x
log tan x
0
- log tan x
0
- (sin x - cos x ) - log tan x
3
0
log tan x
= (– 1) sin x - cos x
log tan x
- log tan x
0
[taking – 1 common from R1, R2 and R3]
= – D [using the reflection property]
fi 2D = 0 fi D = 0.
ax by cz
1
2x
D(x) =
(
y2
1
a b
D2 = x
y
yz z x
(a)
(b)
(c)
(d)
Ans. (d)
(x – 1) (y – 1) (z – 1)
(x – y) (y – z) (z – x)
abc (x – y) (y – z) (z – x)
0
z 2 and
1
c
z , then D1 – D2 equals
xy
)
3 x ( x - 1) x ( x - 1) ( x - 2 ) x x 2 - 1
then D(100) equals
(a) 0
(c) 100!
Ans. (a)
(b) – 100
(d) – 100!
Solution: Taking x common from C2, x + 1 from C3 and
x – 1 from R3, we get
1
1
1
D(x) = x(x + 1) (x – 1) 2 x x - 1 x
3x x - 2 x
Applying C1 Æ C1 – C3, C2 Æ C2 – C3, we get
0 0 1
D(x) = x(x + 1) (x – 1) x -1 x = 0
2 x -2 x
[ C1 and C2 are proportional]
D(100) = 0
Thus,
1
Example 48: If D(x) =
then D(x) equals
(a) x2
2
Example 46: Let D1 = x 2
1
x +1
x ( x + 1)
x
x ( x - 1)
(c) e x - p x
1
-x 2
1
-x 2
(e x + e ) (p x + p )
2
2
(e x - e- x ) (p x - p - x )
2
-2
(b) x2 – 1
2
(d) 0
Ans. (d)
Solution: Using R2 Æ R2 – R3 and (ax + a–x )2 – (ax – a–x )2
= 4ax a–x = 4, we get
D(x) =
1
4
1
4
2
1
4 =0
2
(e x - e- x ) (p x - p - x )
-2
[since R1 and R2 are proportional]
4.16
Complete Mathematics—JEE Main
1
cos x
1 - cos x
Example 49: If D(x) = 1 + sin x cos x 1 + sin x - cos x ,
sin x
sin x
1
p /2
D( x ) dx equals
then Ú
0
(a) 1/4
(c) 0
Ans. (d)
(b) 1/2
(d) – 1/2
Solution: Using C1 Æ C1 – C2 – C3, we get
cos x
1 - cos x
cos x 1 + sin x - cos x
= (– 1) cos x [1 + sin x – cos x – 1 + cos x]
1
= – cos x sin x = - (sin 2x)
2
= (– 1)
p /2
Ú0
p /2
D(x) dx = =
1 ( -1)
˘
cos 2x ˙
2 2
˚0
1
1
(cos p – cos 0) = 4
2
Example 50: The determinant
a
b
aa + b
D=
b
c
ba + c equals zero, if
a a + b ba + c
0
(a)
(b)
(c)
(d)
Ans. (c)
Solution: Applying R2 Æ R2 + R3, and using
sin (A + B) + sin(A – B) = 2 sin A cos B and
cos (A + B) + cos (A – B) = 2 cos A cos B, we get
sin q
cos q
D = 2 sin q cos (2p / 3) 2 cos q cos (2p / 3)
sin (q - 2p / 3)
cos (q - 2p / 3)
0 cos x
1 - cos x
0 cos x 1 + sin x - cos x
-1 sin x
1
D(x) =
\
(a) – sin q – cos q
(b) sin 2q
(c) 1 + sin 2q – cos 2q (d) 0
Ans. (d)
a, b, c are in A.P.
a, b, c are in H.P.
x – a is a factor of ax2 + 2bx + c
x – a is a factor of ax2 + bx + c
sin 2q
2 sin 2q cos ( 4p / 3)
sin (2q - 4p / 3)
sin q
cos q
sin 2q
D =
- sin q
- cos q
- sin 2q
sin (q - 2p / 3) cos (q - 2p / 3) sin (2q - 4p /3)
since
and
fi
cos (2p/3) = cos (p – p/3) = – cos (p/3) = – 1/2
cos (4p/3) = cos (p + p/3) = – cos (p/3) = – 1/2
D=0
[ R1 and R2 are proportional.]
Example 52: If a, b, g are the roots of x3 + px2 + q = 0,
where q π 0, and
1a 1b 1g
D= 1 b 1 g 1 a
1g 1a 1b
then D equals
(a) – p/q
(b) 1/q
(d) 0
(c) p2/q
Ans. (d)
Solution: We have bg + ga + ab = 0. We can write D as
D =
Solution: Applying C3 Æ C3 – aC1 – C2, we get
D =
a
b
b
c
0
0
(
2
a a + b ba + c - aa + 2ba + c
=
)
a b
[expanding along C3]
b c
= – (aa2 + 2ba + c) (ac – b2)
= 0 if x – a is a factor of ax2 + 2bx + c.
1
a 3 b 3g 3
1
a 3 b 3g 3
= – (aa2 + 2ba + c)
Example 51: If q e R, then the determinant
sin q
cos q
sin 2q
D = sin (q + 2p / 3) cos (q + 2p / 3) sin (2q + 4p / 3)
sin (q - 2p / 3) cos (q - 2p / 3) sin (2q - 4p / 3)
equals
=
1
3 3 3
a b g
bg ga ab
ga ab bg
ab bg ga
bg + ga + ab ga ab
ga + ab + bg ab bg
ab + bg + ga bg ga
[using C1 Æ C1 + C2 + C3]
0 ga ab
0 ab bg = 0 [all zero property]
0 bg ga
Example 53: If A, B, C are the angles of a triangle, and
-1 cos C cos B
-1 cos A
D = cos C
cos B cos A
-1
then D equals
Determinants 4.17
(a) 0
(c) 2 cos A cos B cos C
Ans. (a)
(b) – 1
(d) none of these
- a + b cos C + c cos B cos C cos B
1
-1 cos A
a cos C - b + c cos A
D=
a
a cos B + b cos A - c coss A
-1
[using projection formulae and all zero property.]
Example 54: If a, b, c are three complex numbers such
that a2 + b2 + c2 = 0 and
ab
2
ab
c +a
ac
bc
then the value of k is
(a) 1
(c) – 2
Ans. (d)
ac
2
bc
D=
ab
-b
2
ac
bc
ac
-a a
a
bc = abc b -b b
c
c -c
-c2
[taking a, b, c common from C1, C2, C3 respectively]
-1 1 1
2 2 2
= a b c 1 -1 1
1 1 -1
[taking a, b, c common from R1, R2, R3 respectively]
0 2 1
2 1
= 4a2 b2 c2
= a2 b2 c2 2 0 1 = – 2a2 b2 c2
0
1
0 0 -1
Thus,
Example 55: If q, f e R, then the determinant
- sin q
cos q
1
D=
sin q
cos q
1
cos (q + f ) - sin (q + f ) 0
lies in the interval
(b) [– 1, 1]
(c) ÈÎ- 2 , 1˘˚
(d) ÈÎ-1, 2 ˘˚
}
2
b+c a-b a
Example 56: If D1 = c + a b - c b and
a+b c-a c
a b c
D2 = b c a , then D1 – D2 equal
c a b
(a) 0
(c) 6abc
Ans. (a)
(b) 3abc
(d) 2(a3 + b3 + c3)
Solution: Using C1 Æ C1 + C3 and C2 Æ C2
D1, we get
a + b + c -b a
D1 = a + b + c - c b
a + b + c -a c
Using C1 Æ C1 + C2 – C3, we get
c -b a
a b
a -b c
D1 = a - c b = - b - c a = b c
b -a c
c a
c -a b
– C3 in
c
a
b
fi D1 = D2 fi D1 – D2 = 0.
Example 57: If x, y, z are different from zero and
a
b-y c-z
b
c - z = 0,
D= a- x
a-x b-y
c
[applying C1 Æ C1 + C3 and C3 Æ C2 + C3]
k = 4.
(a) ÈÎ- 2 , 2 ˘˚
1
1
sin j - cos f
As – 1 £ sin (f – p/4) £ 1, - 2 £ 2 sin (j – p/4) £
or - 2 £ D £ 2 .
a 2 + b2
Solution: Using a2 + b2 + c2 = 0, we can write D as
ab
- sin q
cos q
0
= (sin j – cos j) (cos2 q + sin2 q)
1
1
= 2
sin j cos j = 2 sin (f – p/4)
2
2
= ka2 b2 c2,
(b) 2
(d) 4
-a2
cos q
D = sin q
0
{
0 cos C cos B
1
-1 cos A = 0
=
0
a
-1
0 cos A
D=
Solution: Applying R3 Æ R3 – cos j R1 + sin j R2, we
get
Solution: Applying C1 Æ aC1 + bC2 + cC3, we get
b2 + c2
Ans. (a)
a b c
+ + is
x y z
(b) –1
(d) 2
then the value of the expression
(a) 0
(c) 1
Ans. (d)
Solution: Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get
a b-y c-z
D = -x
y
0
=0
-x
0
z
4.18
Complete Mathematics—JEE Main
Expanding along C3, we get
-x y
a b-y
+z
=0
-x 0
-x
y
(c – z) (xy) + z(ay + bx – xy) = 0
cxy – xyz + ayz + bxz – xyz = 0
ayz + bzx + cxy = 2xyz
a b c
+ + =2
x y z
(c – z)
fi
fi
fi
fi
Example 58: If p + q + r = a + b + c = 0, then the
determinant
pa qb rc
D = qc
rb
ra pb equals
pc qa
(a) 0
(c) pa + qb + rc
Ans. (a)
(b) 1
(d) none of these
Solution: We have
D = pqr (a3 + b3 + c3) – abc (p3 + q3 + r3)
But a + b + c = 0 fi (a + b)3 = – c3
fi
a3 + b3 + 3ab(a + b) + c3 = 0
fi
a3 + b3 + c3 = – 3ab(– c) = 3abc
Similarly, p3 + q3 + r3 = 3pqr
Thus, D = pqr (3abc) – abc (3pqr) = 0
Example 59: The system of equations
lx + y + z = 0
– x + ly + z = 0
– x – y + lz = 0
will have a non-trivial solution if real values of l are
(a) 0, 1
(b) 0, – 1
(c) 0, 2
(d) 0
Ans. (d)
Solution: The system of equation will have a non-trivial solution if and only if
D = 0 where
l
1 1
D = - 1 l 1 = l3 + 3l = l (l2 + 1)
-1 -1 l
Thus, D = 0, l ΠR fi l = 0
Example 60: The values of k for which the system of
equations
x + ky – 3z = 0
(1)
3x + ky – 2z = 0
(2)
2x + 3y – 4z = 0
(3)
has a non-trivial solution is (are)
21
10
(c) – 5
Ans. (a)
31
10
(d) 4
(b)
(a)
Solution:
From (1) and (2), 2x + z = 0 fi 2x = – z.
From (3), we get 3y = 5z
1
5
z, y = z
2
3
Substituting this in (1) we get
\
x= –
–
Now,
1
5k
z+
z – 3z = 0. fi Ê 5k - 7 ˆ z = 0
Ë 3 2¯
2
3
zπ0 fi
k=
21
.
10
Example 61: If a ΠR and the system of equations x +
ay = 0, az + y = 0, ax + z = 0 has infinite number of solution
then the value of a is
(a) – 1
(b) 1
(c) 0
(d) no real value
Ans. (a)
Solution: We have x = – ay, therefore
– a2y + z = 0 fi z = a2y.
Thus,
az + y = 0 fi (a3 + 1)z = 0.
For the system of equations to have infinite number of solutions a3 + 1 = 0
fi
a=–1
[\ a is real]
Example 62: Given 2x – y + 2z = 2,
x – 2y + z = – 4,
x + y + lz = 4
then the value of l such that the given system of equations
has no solution, is
(a) 3
(b) 1
(c) 0
(d) – 3
Ans. (b)
Solution: From the first two equations, we get
x + y= 6 – z
Putting this value in the last equation we get
6 + (– 1 + l) z = 4 = (l – 1)z = – 2.
For the system of equations to have no solutions, l = 1
Example 63: If the system of linear equations
x + 2ay + az = 0
x + 3by + bz = 0
x + 4cy + cz = 0
has a non-zero solution, then a, b, c
(a) are in G.P.
(b) are in H.P.
Determinants 4.19
(c) satisfy a + 2b + 3c = 0
(d) are in A.P.
Ans. (b)
Solution: The system will have a non-zero solution if
1 2a a
D = 1 3b b = 0
1 4c
c
Using C2 Æ C2 – 2C3, we get
1 0 a
D = 1 b b =0
1 2c c
fi 1(bc – 2bc) + a(2c – b) = 0 fi 2ac = bc + ab
fi
2ac
= b fi a, b, c are in H.P.
a+c
Example 64: The system of homogenous equations
(a – 1)x + (a + 2)y + az = 0
(a + 1)x + ay + (a + 2)z = 0
ax + (a + 1)y + (a – 1)z = 0
has a non-trivial solution if a equals
1
1
(a)
(b) –
2
2
(c) 2
(d) – 1.
Ans. (b)
Solution: The system of equations will have a nontrivial solution if D = 0 where
a -1 a + 2
a
D = a +1
a
a+2
a
a +1 a -1
Using C1 Æ C1 – C3 and C2 Æ C2 – C3, we obtain
-1 2
a
D = -1 - 2 a + 2
1
2 a -1
Using R3 Æ R3 + R2, we get
-1 2
a
D = - 1 - 2 a + 2 = (2a + 1) (2 + 2) = 4(2a + 1)
0
0 2a + 1
Now, D = 0 fi a = – 1/2.
Example 65: The system of equations
ax + y + z = a – 1
x + ay + z = a – 1
x + y + az = a – 1
has no solution, if a equals
(a) – 2
(c) – 2
Ans. (a)
(b) 1
(d) either – 2 or 1
Solution: Let
a 1 1
D= 1 a 1
1 1 a
Using C1 Æ C1 + C2 + C3, we get
1 1 1
1
1
1
D = (a + 2) 1 a 1 = (a + 2) 0 a - 1
0
1 1 a
0
1
a -1
[using R2 Æ R2 – R1, R3 Æ R3 – R1]
\
D = (a + 2) (a – 1)2
For no solution, D = 0 fi a = – 2, 1.
For a = 1, the system of equations has infinite number of
solutions.
For a = – 2, on adding the three equations we obtain
0= – 9
Thus, system of equations has no solution for a = – 2.
Example 66: Suppose a, b, c, a Œ R and abc a π 0. If
the system of equations:
(a + a)x + ay + az = 0
(1)
ax + (b + a)y + az = 0
(2)
ax + ay + (a + c)z = 0
(3)
Ê 1 1 1ˆ
has a non-trivial solution, then a Ë + + ¯ is equal to
a b c
(a) – 1
(b) 0
(c) abc
(d) bc + ca + ab
Ans. (a)
Solution: From (1) and (2)
ax – by = 0
and from (2) and (3)
ax – cz = 0
\
ax = by = cz
x
y
z
=
=
fi
1/ a
1/ b
1/ c
Putting in (1) we get
a +a a a
+ + =0
a
b c
1
1
1ˆ
Ê
fi a Ë + + ¯ = –1
a b c
Example 67: Suppose a, b, Œ R and a, b π 1. If the
system of equations:
ax + y + z = 0
(1)
x + by + z = 0
(2)
x + y + 2z = 0
(3)
has a non-trivial solution, then
4.20
Complete Mathematics—JEE Main
(a) a + b = 2
1
(c) a + = 2
b
Ans. (a)
p
p
(c) - 2 15 2
Ans. (d)
(d) a + b = 0
sec x
= cos x –
Thus,
2
2
1 + b2 x
(1 + c 2 ) x
(1 + a 2 ) x (1 + b2 ) x
1 + c2 x
p /2
Úp / 4
Solution: Applying C1 Æ C1 + C2 + C3, we get
1 (1 + b2 ) x (1 + c 2 ) x
2
cos2 x
cosec 2 x
cos3 x
sin x
Ê 1
ˆ
1
- sin x˜ cos x
(1 – cos 2x) – Á
Ë sin x
¯
2
f(x) dx =
p /2
Úp / 4
cosx dx –
(1 + c ) x
2
1 + c2 x
1 (1 + b ) x
[using a2 + b2 + c2 = – 2]
Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get
1 (1 + b2 ) x (1 + c 2 ) x
f(x) = 0
= (1 – x)2
1- x
0
0
0
1- x
which is a polynomial of degree 2.
Example 69:
sec x
cos x
2
2
f(x) = cos x cos x
2
cos x
f(x)dx is
1
2
p /2
Úp / 4
= 1-
cos 2x dx –
p /2
Úp / 4
1
2
p /2
Úp / 4
dx
Ê 1
ˆ
ÁË sin x - sin x˜¯ cos xdx
1
Ê
1Êp pˆ 1
t2 ˆ ˘
- Á - ˜ + (0 - 1) - Á log t - ˜ ˙
2¯ ˚1
Ë
2 2 Ë 2 4¯ 4
1
1
2
-
2
where t = sin x
p 1
- log 2
8 2
Example 70: If a + b + c = 0, then a root of the equation
a-x
c
b
D=
= 0 is
c
b-x
a
2
1+ b x
+
= 1-
then f(x) is a polynomial of degree
(a) 3
(b) 2
(c) 1
(d) 0
Ans. (b)
p /2
sec 2 x + cot x cosec 2 x
(1 + b ) x (1 + c ) x
f(x) = (1 + a 2 ) x
Úp / 4
cosec 2 x
cos x
= cos x – sin2 x –
1+ a x
then value of
0
= sin2 x [cos x cosec2 x
– cos2 x (sec2 x + cot x cosec2 x)]
2
1
0
cos x
= – (– sin2 x)
Example 68: If a2 + b2 + c2 = – 2 and
Let
sec 2 x + cot x cosec 2 x
2
1
1
a
=1- b
1- a
1 – b = –1 + a
a + b =2
f(x) = 1
cos x
f(x) = - sin 2 x
Putting in (1) we get
a
1
+
+1 = 0
1- a 1- b
fi
fi
(d) none of these
Solution: Applying R2 Æ R2 – R3, we get
Solution: From (1) and (3)
(a – 1)x – z = 0
and from (2) and (3)
(b – 1)y = z
x
y
z
\
=
=
1
1
1
(1 - a )
(1 - b)
fi
(b) p/48
(a) 0
(b) a + b = ab
sec 2 x + cot x cosec 2 x
2
cosec x
2
cosec x
b
(a) 1
(c) a2 + b2 + c2
Ans. (d)
a
c-x
(b) – 1
(d) 0
Solution: Applying C1 Æ C1 + C2 + C3, we get
a+b+c-x
c
b
D = a+b+c-x b-x
a
a+b+c-x
a
c-x
-x
c
b
= -x b - x
a
-x
a
c-x
[
a + b + c = 0]
D clearly equals 0 when x = 0.
It is unnecessary to evaluate the determinant further.
Determinants 4.21
Example 71: A root of the equation
0
D= x+a
x+b
x-a
0
x+c
1
(a)
(a + b + c)
2
(c) –1
Ans. (b)
cos2 A + sin 2 A
x-b
x - c = 0 is
0
cos2 B + sin 2 B
cos C cos A + sin C sin A cos B cos C + sin B sin C
= cos A cos B + sin A sin B
(b) 0
cos C cos A + sin A sin C
cos B cos C + sin B sin C
(d) 1
cos2 C + sin 2 C
Solution: When we substitute x = 0, D becomes
0 - a -b
a 0 -c
b c
0
Alternative Solution
Evaluating D along R1, we get
x+a
x+b
x-c
x+a
+ (x – b)
0
x+b
0
=0
x+c
fi
(x – a) (x + b) (x – c) + (x – b) (x + a) (x + c) = 0
fi
x3 + (– a + b – c)x2 + (– ab + ac – bc)x + abc + x3
+ (a – b + c)x2 + (– ab + ac – bc) – abc = 0
fi
2x3 – 2(ab – ac + bc)x = 0
fi
2x[x2 – (ab – ac + bc)] = 0
fi x = 0 or x = ± ab - ac + bc
\ one of the roots of the equation is 0.
Example 72: If a, b, g are three real numbers such that
a + b + g = 0, then
1
D = cos g
cos b
(a) – 1
(c) 1
Ans. (b)
cos A sin A 0 cos A sin A 0
= cos B sin B 0 cos B sin B 0 = (0) (0) = 0.
cos C sin C 0 cos C sin C 0
Example 73: If a, b, c are the sides of a D ABC opposite
angles A, B, C respectively, and
which is equal to 0 as D is skew symmetric determinant of
odd order.
– (x – a)
cos A cos B + sin A sin B
cos g
1
cos a
cos b
cos a equals
1
a2
b sin A
c sin A
D = b sin A
1
cos ( B - C ) , then D equals
1
c sin A cos ( B - C )
(a) sin A – sin C sin B
(c) 1
Ans. (d)
Solution: By the law of sines
c
a
b
=
=
= k (say)
sin C
sin A
sin B
fi a = k sin A, b = k sin B, c = k sin C. Now
a2
ab / k
ac / k
1
cos ( B - C )
D = ab / k
ac / k cos ( B - C )
1
=a
2
=a
2
=a
2
(b) 0
(d) cos a cos b cos g
Solution: Let A, B and C be three real numbers such
that a = B – C, b = C – A and g = A – B, clearly a + b + g
= 0. We have
1
cos ( A - B ) cos (C - A)
1
cos ( B - C )
D = cos ( A - B )
cos (C - A) cos ( B - C )
1
(b) abc
(d) 0
1
sin B
sin C
sin B
1
cos ( B - C )
sin C cos ( B - C )
1
1
sin ( A + C ) sin ( A + B )
sin ( A + C )
1
cos ( B - C )
sin ( A + B ) cos ( B - C )
1
sin A cos A 0
cos C sin C 0
cos B sin B 0
sin A cos A 0
cos C sin C 0
cos B sin B 0
= a2(0) = 0
Example 74: If a, b, c are distinct, and
= (b – c) (c – a) (a – b) (a + b + c)
1
a
1
b
1
c
a3
b3
c3
4.22
Complete Mathematics—JEE Main
1
then D =
1
1
( x - a )2
( x - b)2
( x - c )2
( x - b) ( x - c) ( x - c) ( x - a ) ( x - a ) ( x - b)
D=
vanishes if
1
(a + b + c)
3
2
(a + b + c)
(b) x =
3
(c) x = a + b + c
(d) none of these
Ans. (a)
(a) x =
B
C
3
3
D= A
1
B
1
C
1
(a) x =
= (– 1) (– 1)
1
A
1
B
b
g
a2
b2
g2
equals
25 d
2a
(b)
(c)
-25d
a+b+c+d
(d) none of these
D = abg
1
C
A3
B3 C 3
x-b
x-c
x-a
x-c
x-a =0
x-b
1
1
(a + b + c) (b) x =
(a + b + c)
3
2
(c) x = a + b + c
Ans. (a)
a
25 d
a
Solution: Taking a, b, g common from C1, C2, C3
respectively, we get
= abg
(d) x = 0
Solution: Using C1 Æ C1 + C2 + C3, we get
3 x - (a + b + c ) x - b
D = 3 x - (a + b + c ) x - c
3 x - (a + b + c ) x - a
Note that D becomes 0 when x =
x-c
x-a
x-b
1
(a + b + c).
3
Example 76: If a, b, g are different from 1 and are the
roots of ax3 + bx2 + cx + d = 0 and (b – g) (g – a) (a – b) =
25/2, then the determinant
1
1-a
1
1- b
1
1- g
1
a
1
b
1
g
1
1-a
1
1
1- b 1-a
1
1
1- g 1-a
1
a
0
b -a
0
g -a
[using C2 Æ C2 – C1, and C3 Æ C3 – C1]
= (B – C) (C – A) (A – B) (A + B + C)
= (c – b) (a – b) (b – a) [3x – (a + b + c)]
1
Note that D become 0 when x = (a + b + c)
3
Example 75: The equation
x-a
D= x-b
x-c
is satisfied when
g
1- g
(a)
where A = x – a, B = x – b, C = x – c.
3
b
1- b
Ans. (d)
Solution: Multiplying C1 by (x – a), C2 by (x – b) and C3 by
(x – c), we get
A
B
C
1
A3
B3
C3
D=
ABC
ABC ABC ABC
A
a
1-a
=
abg ( -1) ( b - a ) (g - a ) 1 - g
1
(1 - a ) (1 - b ) (1 - g )
=
abg (a - b ) ( b - g ) (g - a )
(1 - a ) (1 - b ) (1 - g )
1- b
1
As a, b, g are the roots of ax3 + bx2 + cx + d = 0,
ax3 + bx2 + cx + d = a(x – a) (x – b) (x – g)
and
abg = – d/a
D=
Thus,
25 d
(- d / a )(25 / 2 )
= 2 (a + b + c + d )
(a + b + c + d ) / a
Example 77: Let P = [aij] be a 3 ¥ 3 matrix and Q = [bij],
where bij = 2 i + j aij for l £ i, j £ 3. If the determinant of P is
2, then the determinant of the matrix Q is
(a) 210
(b) 211
(c) 212
Ans. (d)
(d) 213
Solution:
22 a11
23 a12
2 4 a13
det (Q)= 23 a21
2 4 a22
25 a23
2 4 a31
25 a32
26 a33
Determinants 4.23
a11
2
3
4
= (2 )(2 )(2 ) 2a21
a12
2a22
a13
2a23
22 a31 22 a32
22 a33
a11
= 2 (2) (2 ) a21
a31
9
2
a12
a22
a32
a + 1 a - 1 (- 1)n + 2 a
= b + 1 b - 1 (- 1)n + 1 b
a + (- 1)n + 2 a
a +1 a -1
n +1
b b +1 b -1
c -1 c +1
b +1
b -1
Example 80: Let w be the complex number cos
+ i sin
c -1
c +1 = 0
z +1
w
w2
w
z + w2
1
w2
1
z +w
Solution: Denote the given determinant by D.
Using C1 Æ C1 + C2 + C3 and 1 + w + w2 = 0, we get
1
D = z 1 z + w2
1
1
a (- 1)
n +1
1
z +w
Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get
0
b +1
b -1
w2
w
w
D = z 0 z + w2 - w
Then the value of n is
(a) any odd integer
(b) any integer
(c) zero
(d) any even integer
Ans. (a)
c -1
c +1
n
b (- 1) c
= 0 is
(b) 0
(d) 3
1
(-1)n + 2 a (-1)n + 1 b (-1)n c
2p
3
2p
. Then the number of distinct complex number z
3
(a) 1
(c) 2
Ans. (a)
Example 79: Let a, b, c be such that b(a + c) π 0.
a a +1 a -1
If D = - b b + 1 b - 1
c c -1 c +1
(- 1)
c -1 c +1
Thus,
satisfying
1 x + 1 ( x + 1) ( x + 2)
D1(x) = 0
=2
1
2 ( x + 2)
0
1
2 ( x + 3)
n+2
(- 1) c
The first column consists of all 0’s if n is any odd integer.
\ D = 0 if n is any odd integer.
Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get
a +1
a -1
b +1 b -1
n
c + (- 1) c
1 x + 1 ( x + 1) ( x + 2)
D1(x) = 1 x + 2 ( x + 2) ( x + 3)
1 x + 3 ( x + 3) ( x + 4)
Solution: Let D2 =
= (–1)2 (- 1)n + 1 b
n
Solution: Taking x! common from R1, (x + 1)! from R2
and (x + 2)! from R3 to obtain
D(x) = x! (x + 1)! (x + 2)! D1(x) where
+
(- 1)n + 2 a a + 1 a - 1
D = - b + (- 1)
2x! (x + 1)!
2x! (x + 1)! (x + 2)!
2x! (x + 3)!
2(x + 1)! (x + 2)! (x + 3)!
a +1
a -1
[using the reflection property]
a13
a23 = 212 det (P) = 213
a33
Example 78: If x is a positive integer, and D(x) =
x!
( x + 1)! ( x + 2)!
( x + 1)! ( x + 2)! ( x + 3)! , then D(x) is equal to
( x + 2)! ( x + 3)! ( x + 4)!
(a)
(b)
(b)
(d)
Ans. (b)
(- 1)n c
c -1 c +1
1-w
w2
1 - w2
z + w - w2
= z[(z + w2 – w) (z + w – w2) – (1 – w) (1 – w2)]
= z[z2 – (w2 – w)2 – (1 – w – w2 + 1)]
= z[z2 – (w4 + w2 – 2w3) – 3] = z(z2) = z3
3
Thus, z = 0 fi z = 0
Thus, there is just one value of z, satisfying the given
equation.
4.24
Complete Mathematics—JEE Main
Assertion-Reason Type Questions
an
an + 3
an + 6
Example 81: Let Dn = an + 1
an + 4
an + 7
an + 2
an + 5
an + 8
Example 83: Suppose x > 0, y > 0, z > 0 and
x log 2 3 15 + log (a x )
D(a, b, c) = y log 3 5 25 + log (b y )
Statement-1: If ak > 0 " k ≥ 1 and a1, a2, a3, . . . are in G.P.
then
Dn = 0 " n ≥ 1.
Statement-2: If a1, a2, a3 . . . are in A.P. then
Dn = 0 " n ≥ 1.
Ans. (b)
Solution: Let ak = ar
k–1
" k ≥ 1, then
D n = a na n + 3 a n + 6
1
r
1
r
1
r
2
2
2
=0
r r r
Next, if ak = b + (k – 1)d, then using C2 Æ C3 – C1 and
C2 Æ C2 – C1, we get
an
Dn = an + 1
an + 2
Statement-1: D(8, 27, 125) = 0
Ê 1 1 1ˆ
Statement-2: D Á , , ˜ = 0
Ë 2 3 5¯
Ans. (b)
Solution: Using log (bc) =
C3 – 5C2 we get
x log 2 3
D(a, b, c) = y log 3 5
z log 5 7
x log a
y log b
z log c
Example 84: Let a π 0, p π 0 and
a b c
3d 3d
C2 and C3 are identical]
0
cos x - sin x
0
cos x
Example 82: Let f(x) = sin x
cos x sin x
0
Statement-1: If sin 2x = 1, then f(x) = 2/3
Statement-2: f(x) = 0 if sin x = cos x
Ans. (d)
c log b and applying C3 Æ
D(8, 27, 125) = D(23, 33, 53) = 0 as in this case C1 and C3 are
proportional. Similarly, D(1/2, 1/3, 1/5) = D(2–1, 3–1, 5–1) = 0.
3d 3d
3 d 3d = 0
[
2
.
Solution: Multiplication of two determinants leads us
1 -y y
f(x) = - y 1 y
y
y 1
where
y = sin x cos x
Using C1 Æ C1 – C2, C2 Æ C2 + C3, we get
f(x) = (1 + y)2
z log 5 7 35 + log (c z )
1 0 y
1 0 y
2
-1 1 y = (1 + y) 0 1 2 y
0 1 1
0 1 1
= (1 + y)2 (1 – 2y)
When sin 2x = 1, y = 1/2 and f(x) = 0
When sin x = cos x, 1 – 2y = 1 – 2 sin2x = cos 2x = 0
\
f(x) = 0.
D= 0 p q
p q 0
Statement-1: If the equations
ax2 + bx + c = 0 and px + q = 0
have a common root, then D = 0.
Statement-2: If D = 0, then the equations
ax2 + bx + c = 0 and px + q = 0
have a common root.
Ans. (b)
Solution: If l is a common root of
ax2 + bx + c = 0 and px + q = 0, then
al2 + bl + c = 0, pl + q = 0 and pl2 + ql = 0
Eliminating l, we obtained D = 0.
For statement-2, expanding D along C1 we obtain
– aq2 + p(bq – cp) = 0
2
or
Ê qˆ
Ê qˆ
aÁ - ˜ + bÁ - ˜ + c = 0
Ë p¯
Ë p¯
Thus, ax2 + bx + c = 0 and px + q = 0 have a common root.
Example 85: Statement-1:
sin p
cos( x + p / 4) tan( x - p / 4)
log( x / y)
=0
D = sin( x - p / 4) - cos(p / 2)
cot(p / 4 + x )
log( y / x )
tan p
Determinants 4.25
Statement-2: A skew symmetric determinant of odd order
equals 0.
Ans. (a)
Solution: For statement-2, see theory.
Now, using
pˆ
Èp Ê p
ˆ˘
Ê
cos Á x + ˜ = cos Í - Á - x˜ ˙
Ë
¯˚
Ë
4¯
2
4
Î
pˆ
Êp
ˆ
Ê
= sin Á - x˜ = - sin Á x - ˜ ;
Ë4
¯
Ë
4¯
0 -1 3
Solution: e = f (0) = 1 0 - 4
-3 4
0
=0
[skew symmetric determinant of odd order]
\ Statement-2 is true.
1
To obtain a, replace x by , so that
x
2
3
Ê 1ˆ
Ë x¯ + x
1
+1
x
1
-3
x
Êp
ˆ
Èp Ê p
ˆ˘
cot Á + x˜ = cot Í - Á - x˜ ˙
Ë4
¯
Ë
¯˚
4
Î2
p
p
= tan ÊÁ - xˆ˜ = - tan ÊÁ x - ˆ˜ ,
Ë4
¯
Ë
4¯
and
log (x/y) = – log (y/x)
we find D is a skew symmetric determinant of odd order.
Example 86: Let f : Q Æ [–1, 1] by
f (x) = sin x
Let x1, x2, x3 be three distinct rational numbers. Let a = f(x1),
b = f(x2), c = f(x3).
Statement-1:
1 a a2
D= 1 b b
2
π0
1 c c
Statement-2: f is a one-to-one function.
Ans. (a)
Solution: Let x1, x2 ΠQ be such that
f(x1) = f(x2)
fi
sin x1 = sin x2
fi
x1 = np + (–1)n x2, n Œ I
fi
x1 + (–1)n+1 x2 = np, n Œ I.
But LHS is rational and RHS is irrational except when n = 0.
\
x1 = x2
Thus, f is one-to-one.
Also,
D = (a – b) (b – c) (c – a) π 0
as x1, x2, x3 are distinct and hence a, b, c are distinct.
Example 87: Let
x2 + 3x
f (x) = x + 1
x-3
x -1 x + 3
-2 x x - 4
x + 4 3x
= ax4 + bx3 + cx2 + dx + e
Statement-1: a = – 1
Statement-2: e = f (0)
Ans. (a)
1
+3
x
1
-4
x
3
x
4
3
2
Ê 1ˆ
Ê 1ˆ
Ê 1ˆ
Ê 1ˆ
aË ¯ + bË ¯ + cË ¯ + d Ë ¯ + e
x
x
x
x
2
Take x common from C1, x from C2 and C3
1 + 3x 1 - x 1 + 3x
1
1 - 4x
= 4 x + x2
-2
x
2
x - 3x 1 + 4 x
3x
1
[a + bx + cx 2 + dx 3 + ex 4 ]
x4
Cancel 1/x4 and put x = 0 to obtain
1 1 1
0 -2 1 = a
0 1 0
fi
a = –1
=
2
1
-1
x
2
x
1
+4
x
Example 88: Statement-1: The system of line equations
x + (sin a) y + (cos a) z = 0
x + (cos a) y + (sin a) z = 0
x – (sin a) y – (cos a) z = 0
has a non-trivial solution for only one value of a lying in
the interval (0, p /2).
Statement-2: The equation in a
cos a sin a
cos a
D = sin a cos a
sin a = 0
cos a - sin a - cos a
has only one solution lying in the interval (0, p /2).
Ans. (b)
Solution: Using C1 Æ C1 – C3 in D, we get
0
0
D=
2 cos a
sin a
cos a
- sin a
cos a
sin a
- cos a
= 2 cos a (sin2 a – cos2 a)
4.26
Complete Mathematics—JEE Main
= –2 cos a cos 2 a
D = 0 for a = p/4 Π(0, p /2).
This is the only value of a lying in (0, p /2) for which D = 0.
The system of linear equations will have a non-trivial solution if and only if
1 sin a
cos a
sin a = 0
D1 = 1 cos a
1 - sin a - cos a
Using R2 Æ R2 + R1, we get
2
0
D1 = 1 cos a
1 - sin a
0
sin a = 0
- cos a
fi
2 (– cos2 a + sin2 a) = 0
fi
–2 cos 2a = 0
This is true for only one value of a Π(0, p /2)
viz, a = p /4.
Thus, statement-1 is also true. However, statement-2 is not
a correct reason for statement-1.
LEVEL 2
Straight Objective Type Questions
Example 89: Consider the system of equations in x, y, z:
(sin 3q) x – y + z = 0
(cos 2q) x + 4y + 3z = 0
2x + 7y + 7z = 0
The values of q for which the system has a non-trivial solution are given by
1
Ê
ˆ
(a) p Á n + (- 1)n ˜
Ë
¯
3
1
(b) p ÊÁ n + (- 1)n ˆ˜
Ë
¯
4
1
Ê
ˆ
(b) p Á n + (- 1)n ˜
Ë
¯
6
(d)
np
2
Example 90:
D(x) =
Let
then
p /2
Ú- p / 2 x D( x)dx
2 sin 4 x
3 + 2 cos4 x
sin 2 2 x
2 sin 4 x
2 cos4 x
3 + sin 2 2 x
equals
(b) p (p – 1)
(d) 0
1
2 cos4 x
D(x) = f(x) 1 3 + 2 cos4 x
equations to have a non1
3 =0
7
sin 3q
D = cos 2 q + 4 sin 3q
2 + 7 sin 3q
fi
sin 2 2 x
Solution: Using C1 Æ C1 + C2 + C3, we get
Using R2 Æ R2 + 4R1 and R3 Æ R3 + 7R1 we get
fi
fi
fi
fi
fi
2 cos4 x
(a) p2
(c) 1
Ans. (d)
where n is an integer
Ans. (c)
Solution: For the system of
trivial solution we must have
sin 3q - 1
D = cos 2q 4
2
7
3 + 2 sin 4 x
-1 1
0 7 =0
0 14
2(cos 2q + 4 sin 3q) – (2 + 7 sin 3q) = 0
sin 3q + 2 cos 2q – 2 = 0
3 sin q – 4 sin3 q – 4 sin2 q = 0
sin q(2 sin q + 3) (2 sin q – 1) = 0
sin q = 0 or sin q = 1/2
[ |sin q | £ 1]
Ê
(- 1)n ˆ
q = m p or q = n Á p +
where m
6 ˜¯
Ë
and n are integers
1
2 cos4 x
sin 2 2 x
sin 2 2 x
3 + sin 2 2 x
f(x) = 3 + 2 sin4x + 2 cos4x + sin2 2x
where
= 3 + 2 (sin4x + cos4x + 2 sin2x cos2 x)
= 3 + 2 (cos2x + sin2x)2 = 5
Applying C2 Æ C2 – (2 cos4 x) C1 and C3 Æ C3 –
(sin2 2x) C1, we get
1 0 0
D(x) = 5 1 3 0 = 45
1 0 3
Thus,
p /2
p /2
Ú- p / 2 x D( x)dx = 45Ú- p / 2 xdx = 0
Example 91: If f (x), g(x)and h(x) are three polynomials
of degree 3 then
f ¢( x )
g¢ (x) h¢ (x)
f(x) =
f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x )
f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x )
Determinants 4.27
is a polynomial of degree
(a) 3
(c) 5
Ans. (d)
(b) 4
(d) none of these
Solution: The function D(x) is continuous on [0, p/2]
and differentiable on (0, p/2). Also D(0) = 0 and D(p/2) = 0.
Thus, by the Rolle’s theorem there exists at least one
c e (0, p/2) such that D¢(c) = 0.
cos2 x
Solution: We have
f ¢¢( x )
f¢(x) =
g ¢¢ ( x ) h ¢¢ ( x )
f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x ) +
f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x )
f ¢( x ) g ¢ ( x ) h ¢ ( x )
f ¢( x )
g¢ (x) h¢ (x)
f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x ) + f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x )
f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x )
f iv ( x ) giv ( x ) hiv ( x )
f ¢( x ) g ¢ ( x ) h ¢ ( x )
= 0 + 0 + f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x ) = 0
0
0
0
since f, g, h are polynomials of degree 3, f iv(x) = giv(x) =
hiv(x) = 0
fi f(x) must be a constant.
equals
(a) 9b2(a + b)
(c) 9(a + b)3
Ans. (a)
(b) 9a2(a + b)
(d) 9ab(a + b)
then
p /2
Ú0
Applying R2 Æ R2 – R1 and R3 Æ R3 – R1,
1 a + b a + 2b
-b
D = 3(a + b) 0 -b
0
b
-2b
-b -b
= 3(a + b)
= 9b2(a + b)
b -2b
Example 93: Let
sin x cos x sin 2 x + cos 2 x
1
1
D(x) = 0
1
0
-1
then D¢(x) vanishes at least once in
(a) (0, p/2)
(b) (p/2, p)
(c) (0, p/4)
(d) (– p/2, 0)
Ans. (a)
cos x
0
[D(x) + D¢(x)] dx equals
(a) p/3
(c) 2p
Ans. (b)
(b) p/2
(d) 3p/2
Solution: Applying C1 Æ C1 – sin x C3 and C2 Æ C2 +
cos x C3, we get
1
0
- sin x
D(x) = 0
1
cos x
0
sin x - cos x
Applying R3 Æ R3 – sin x R1 + cosx R2, we get
- sin x
cos x
1 0
D(x) = 0 1
2
=1
2
0 0 cos x + sin x
fi
D¢(x) = 0
Thus,
p /2
Ú0
[D(x) + D¢(x)] dx =
p /2
Ú0
dx =
p
.
2
Example 95: The determinant
Solution: Applying C1 Æ C1 + C2 + C3, we get
1 a + b a + 2b
D = 3(a + b) 1
a
a+b
1 a + 2b
a
sin 2 x
- cos x
Example 94: Let D(x) = cos x sin x
sin x
Example 92: The determinant
a
a + b a + 2b
a
a+b
D = a + 2b
a + b a + 2b
a
cos x sin x - sin x
D=
13 + 3
2 5
15 + 26
5
10 equals
3 + 65
15
5
(a) 15 2 - 25 3
5
(b) 25 3 - 15 2
(d) -15 2 + 7 3
(c) 3 5
Ans. (a)
Solution: Taking
5 common from C2 and C3, we get
13 + 3
D=
2
( 5)
Applying C1 Æ C1 –
D = ( 5)
= 5
(
1
15 + 26
5
2
3 + 65
3
5
13 C3 –
- 3
2
3 C2, we get
2
1
(
)(
0
5
2 = 5 - 3 5- 6
0
3
5
)
18 - 25 3 = 15 2 - 25 3
)
4.28
Complete Mathematics—JEE Main
Example 96: The values of l for which the system of
equations
x + y – 3= 0
Example 98: If a + b + c π 0, the system of equations
(b + c) (y + z) – ax = b – c
(c + a) (z + x) – by = c – a
(a + b) (x + y) – cz = a – b
(1 + l)x + (2 + l)y – 8 = 0
x – (1 + l)y + (2 + l) = 0
has a non-trivial solution, are
(a) – 5/3, 1
(b) 2/3, – 3
(c) – 1/3, – 3
(d) 0
Ans. (a)
Solution: The given system of equations has a non
trivial solution if
1
1
-3
2+l
-8 = 0
D = 1+ l
1
- (1 + l ) 2 + l
Applying C2 Æ C2 – C1 and C3 Æ C3 + 3C1, we get
1
D = 1+ l
1
fi
fi
fi
fi
0
1
-2 - l
0
-5 + 3l = 0
5+l
(5 + l) + (2 + l) (3l – 5) = 0
5 + l + 6l – 10 + 3l2 – 5l = 0
2
3l + 2l – 5 = 0 fi (3l + 5) (l – 1) = 0
l = – 5/3 or l = 1.
Example 97: The values of m for which the system
of equations 3x + my = m and 2x – 5y = 20 has a solution
satisfying the conditions x > 0, y > 0 are given by the set
(a) {m : m < – 13/2}
(b) {m : m > 17/2}
(c) {m : m < – 13/2 or m > 17/2}
(d) none of these
Ans. (d)
Solution: Here D =
Dx =
3 m
= – 15 – 2m
2 -5
m m
= – 25 m
20 -5
3 m
and Dy =
= 60 – 2m
2 20
If D = 0, then m = – 15/2. But for this value of m, Dx π 0
and Dy π 0. Thus, in this case, the system of equations is not
consistent.
25m
2m - 60
and y =
, by the Cramer’s
If D π 0, then x =
2m + 15
2m + 15
rule.
Now, x > 0, y > 0 ¤ 25m > 0, 2m – 60 > 0, 2m + 15 > 0
or 25m < 0, 2m – 60 < 0, 2m + 15 < 0
fi m > 30 or m < – 15/2.
has
(a)
(b)
(c)
(d)
Ans. (a)
a unique solution
no solution
infinite number of solutions
finitely many solutions
Solution: We can write the above system of equations
(a + b + c) (y + z) – a(x + y + z) = b – c
(a + b + c) (z + x) – b(x + y + z) = c – a
(a + b + c) (x + y) – c(x + y + z) = a – b
Adding the above equations, we obtain
2(a + b + c) (x + y + z) – (a + b + c) (x + y + z) = 0
fi
fi
fi
(a + b + c) (x + y + z) = 0
x+y+z =0
y+z =–x
[
\
(b + c) (– x) – ax = b – c fi x =
c-b
a+b+c
Similarly, y =
a + b + c π 0]
a-c
b-a
,z=
.
a+b+c
a+b+c
Example 99: Let a, b, c be positive real numbers. The
following system of equations in x, y and z.
x2
a2
has
+
y2
b2
-
(a)
(b)
(c)
(d)
Ans. (d)
z2
c2
= 1,
x2
a2
-
y2
b2
+
z2
c2
= 1,
- x2
a2
+
y2
b2
+
z2
c2
=1
no solution
unique solution
infinitely many solutions
finitely many solutions
Solution: Adding all the equations, we obtain
x2
a2
+
y2
b2
+
z2
c2
=3
Subtracting first equation from it we get
2z2
2
= 2 fi z2 = c2
c
fi z = ± c. Similarly, x = ± a, y = ± b. Thus, the given system
of equations has eight solutions.
Example 100: If the system of equations
x – ky – z = 0, kx – y – z = 0, x + y – z = 0
has a non-zero solution, then the possible values of k are
(a) – 1, 2
(b) 1, 2
(c) 0, 1
(d) – 1, 1
Ans. (d)
Determinants 4.29
Solution: As the given system has a non-zero solution,
1 - k -1
1+ k
0 = k -1 -1 = 1 + k
1 1 -1
0
- k - 1 -1
-2
-1
0
-1
[using C1 Æ C1 – C2, C2 Æ C2 + C3]
fi
0 = (– 1) [(1 + k) (– 2) – (1 + k) (– k – 1)]
fi
0 = (1 + k) (– 2 + k + 1) fi k = – 1, 1
Example 101: If the system of equations
lx1 + x2 + x3 = 1, x1 + lx2 + x3 = 1, x1 + x2 + lx3 = 1
is inconsistent, then l equals
(a) 5
(b) – 2/3
(c) – 3
(d) – 2
Ans. (d)
l 1 1
l +2 1 1
Solution: Let D = 1 l 1 = l + 2 l 1
1 1 l
l +2 1 l
[C1 Æ C1 + C2 + C3]
1 1
= (l + 2) 1 l
1 1
1
0
0
1
0
1 = (l + 2) 1 l - 1
1
0
l -1
l
[using C2 Æ C2 – C1 and C3 Æ C3 – C1]
= (l + 2) (l – 1)2
If D = 0, then l = – 2 or l = 1.
But when l = 1, the system of equation becomes x1 + x2 + x3
= 1 which has infinite number of solutions. When l = – 2, by
adding three equations, we obtain 0 = 3 and thus, the system
of equations is inconsistent.
Example 102: If p π a, q π b, r π c and the system of
equations
px + ay + az = 0
bx + qy + bz = 0
cx + cy + rz = 0
has a non-trivial solution, then the value of
p
q
r
+
+
is
p-a q-b r-c
(a) – 1
(c) 1
Ans. (d)
(b) 0
(d) 2
Solution: As the given system of equations has a nontrivial solution
p a a
D= b
c
q b =0
c r
Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we get
p a- p a- p
0
=0
D= b q-b
0
c
r-c
Expanding along C3, we get
p a- p
b q-b
(a – p)
+ (r – c)
=0
b q-b
c
0
fi (a – p) (– c) (q – b) + (r – c) {p(q – b) – b(a – p)} = 0
fi (p – a) (q – b) c + p (r – c) (q – b) + b (r – c) (p – a) = 0
Dividing by (p – a) (q – b) (r – c) we get
c
p
b
+
+
=0
r-c p-a q-b
fi
p
q
r
q-b r-c
+
+
+
=
=2
p-a q-b r-c
q-b r-c
Example 103: Let l and a be real. Relation between l
and a for which the system of equations
lx + (sin a)y + (cos a)z = 0
x + (cos a)y + (sin a)z = 0
– x + (sin a)y – (cos a)z = 0
has a non-trivial solution is
(a) l = sin 2a + cos 2a
(b) l = |sin 2a|
(c) l = |sin 2a – cos 2a|
(d) l = cos 2a
Ans. (a)
Solution: The system of equations will have a nontrivial solution if and only if
l sin a cos a
0 = 1 cos a sin a
-1 sin a - cos a
0 (l + 1)sin a (1 - l )cos a
= 0 cos a + sin a sin a - cos a
-1
sin a
- cos a
[using R1 Æ R1 + lR3, R2 Æ R2 + R3]
= (l + 1) sin a (sin a – cos a)
+ (l – 1) cos a(cos a + sin a)
= l(sin2a + cos2a) + [sin2a – cos2 a
– 2 sin a cos a]
Thus, l = sin 2a + cos 2a
Example 104: If fr(x), gr(x), hr(x), r = 1, 2, 3 are
polynomials in x such that fr(a) = gr(a) = hr(a), r = 1, 2, 3
f1 ( x ) f2 ( x ) f3 ( x )
F (x) = g1 ( x ) g2 ( x ) g3 ( x )
h1 ( x ) h2 ( x ) h3 ( x )
then F ¢(a) is
(a) – 1
(b) a
(c) 0
(d) none of these
Ans. (c)
and
4.30
Complete Mathematics—JEE Main
Solution: Differentiating w.r.t. x we get F ¢(x)
f1¢( x ) f2¢ ( x ) f3¢ ( x )
f1 ( x ) f2 ( x ) f3 ( x )
= g1 ( x ) g2 ( x ) g3 ( x ) + g1¢ ( x ) g2¢ ( x ) g3¢ ( x )
h1 ( x ) h2 ( x ) h3 ( x )
h1 ( x ) h2 ( x ) h3 ( x )
f1 ( x ) f2 ( x ) f3 ( x )
+ g1 ( x ) g2 ( x ) g3 ( x )
h1¢ ( x ) h2¢ ( x ) h3¢ ( x )
Since fr(a) = gr(a) = hr(a) for r = 1, 2, 3, we get each of the
three determinants on the right side becomes zero when x
is replaced by a. (In each case two rows become identical.)
Thus, F ¢(a) = 0.
1
1
(y – w), z = (y + w), y ≥ 0
7
3
(d) x = 1, y = 0, z = 0, w = 0.
Ans. (d)
(c) x =
Solution: We have
1 – (x + 2y) = z ≥ 0
and
2x – 3y – 2 = – w £ 0.
fi
x + 2y £ 1 and 2x – 3y £ 2, x ≥ 0, y ≥ 0
The only point where the two shaded region intersect is
(1, 0). Thus, x = 1, y = 0 for these values z = 0, w = 0.
1
Example 105: The number of real values of a for which
the system of equations
x + ay – z = 0, 2x – y + az = 0, ax + y + 2z = 0
has a non-trivial solution, is
(a) 3
(b) 1
(c) 0
(d) infinite
Ans. (a)
Solution: Since the given system of equations has a
non-trivial solution,
1 a -1
D = 2 -1 a = 0
a 1 2
Using C1 Æ C1 + C3, C2 Æ C2 + aC3, we get
0
2
D = 2 + a -1 + a
2 + a 1 + 2a
a
2
fi
(2 + a) (1 + 2a + 1 – a2) = 0
Example 106: The solution set of
x + 2y + z = 1
subject
(a) x =
1
(y – w),
7
y ≥ w ≥ 0, z ≥ 0
(b) x =
1
(y + w),
8
z=
2x
y
–3
=2
1
x
–2
3
Fig. 4.1
Example 107: The number of values of k for which the
system of equations
(k + 1)x + 8y = 4k
Solution: Let D =
fi
a = – 2, 1 ± 3 .
Thus, there are three real values of a for which the system
of equations has a non-trivial solution.
2x – 3y + w = 2
x ≥ 0, y ≥ 0, z ≥ 0, w ≥ 0
=1
O
2 + a -1 + a 2
=0
2 + a 1 + 2a
(-1)
2y
has infinitely many solutions is
(a) 0
(b) 1
(c) 2
(d) infinite
Ans. (b)
=0
fi
x+
kx + (k + 3)y = 3k – 1
-1
0
1
2
y
is
1
(y – w), y ≥ w ≥ 0
3
k +1
8
k
k +3
= k2 – 4k + 3 = (k – 1) (k – 3)
If D π 0, the system of equations has a unique solution.
For the system of equations to have an infinite number of
solutions
D = 0 fi k = 3, 1
For k = 3, the system equations becomes
4x + 8y = 12 and 3x + 6y = 8
8
3
Thus, in this case the system of equations has no solution.
For k = 1, the system of equations becomes
2x + 8y = 4 and x + 4y = 2
fi
x + 2y = 3, x + 2y =
fi
x + 4y = 2, x + 4y = 2
This system has infinite number of solutions.
Determinants 4.31
Example 108: Suppose a, b, c ΠR and let
0
a-x b-x
f(x) = - a - x
0
c-x
-b-x -c-x
0
(k – 1)x – 2y = 0.
If k = – 1, we can choose y in infinite number of ways,
corresponding to which we can choose x and z in infinite
number of ways.
If k = 1, then y = 0 and x and hence z can be chosen in infinite
number of ways.
2
Then coefficient of x in f(x) is
(a) – (a + b + c)
(b) a + b + c
(c) 0
(d) ab + bc + ca
Ans. (c)
Example 110: Let a2, a3 ΠR be such that | a2 Рa3 | = 6.
Let
1
a3
a2
f(x) = 1
a3
2a2 - x , x ΠR
1 2a3 - x
a2
Solution: Expanding along C1
a-x b-x
a-x b-x
- (b + x )
-c-x
0
0
c-x
= (a + x) (b – x) (c + x) – (b + x) (a – x) (c – x)
f(x) = (a + x)
= – [(x + a) (x – b) (x + c) + (x – a) (x + b) (x – c)]
3
The maximum value of f (x) is
(a) 6
(b) 9
(c) 12
(d) 36
Ans. (b)
2
= – [2x + x (a – b + c – a + b – c) + ...]
= – 2x3 + 0x2 + ...
\ coefficient of x2 in f(x) is 0.
Solution: Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get
Example 109: If the system of equations
x – ky – z = 0, kx – y – z = 0, x + y – z = 0
has a non-zero solution, then possible values of k are
(a) – 1, 2
(b) 1, 2
(c) 0, 1
(d) – 1, 1
Ans. (d)
Solution: Subtracting the last equation from the first
two equations, we get
– (k + 1)y = 0
1
a3
a2
f(x) = 0
0
a2 - x
0 a3 - x
0
= – (a2 – x) (a3 – x) = – [x2 – (a2 + a3)x + a2a3]
=
1
a + a3 ˆ 2
(a2 – a3)2 – Ê x - 2
£9
Ë
4
2 ¯
f(x) attains maximum value 9 when
x=
1
(a2 + a3).
2
EXERCISES
Concept-based
Straight Objective Type Questions
1. Suppose A = (aij)n × n, where aij Œ R.
If det (adj(A)A–1) = 3, then det (adj(A)) equals:
(a)
3
(c) 3 3
(a) – p3
p+2
p+5
p+8
(d) 0
(b) 3
1
x
x2
(d) 9
D = x2
1
x
x
2
1
2. If a, b, c are in A.P. and p is a real number, and
p+c
D = p+b
p+a
then D equals:
(c) p3 – 2 abc
3. Suppose x π 1 and
p+a
p+b
p+c
(b) p3
x
then D = 0
(a) for exactly two distinct complex numbers
(b) for exactly four distinct complex numbers
(c) for exactly two distinct real numbers
(d) none of these
4.32
Complete Mathematics—JEE Main
10! 11! 12!
D
4. Let D = 11! 12! 13! , then
– 260 equals:
(10!)3
12! 13! 14!
(a) 1
(c) 3
(b) 2
(d) 4
x +1
2
3
x+2
3 ,
P(x) = 1
1
2
x+3
the product of zeros of P(x) is
(b) 6
(d) 12
6. Let A(x1, y1), B(x2, y2) and C(x3, y3) be vertices of
an equilateral triangle whose side is 4 units. Let
x1
D = x2
x3
(b) 128
(d) 256
7. Suppose a, b, c are in A.P. If p, q, r are also in
A.P., then value of
x2 + a
x+p c
D = x2 + b
x+q
b ,
x2 + c
is dependent on
(a) x
(c) p, q, r
x+r
a
(a + 2016)2
D = (a - 2016)2
(b - 2016)2
(c - 2016)2
b2
c2
If D = k (2016)3 (a – b) (b – c)(c – a), then k is
equal to:
(a) – 1
(b) – 4
(c) 4
(d) 1
x -6
-1
11. Let P(x) = 2 -3 x x + 3 ,
-3 2 x x + 2
sum of the zeros of P(x) is
(a) – 6
(b) – 7
(c) 13/5
(d) –12/5
a+b
a+b
a 2 + b2
x n+2
x n +3
D = y n +1
yn+2
y n +3
z n +1 z n + 2 z n +3
If D = (x – y) (y – z)(z – x) x2 y2 z2, then n is equal
to:
(a) –1
(b) 0
(c) 1
(d) 2
13. Suppose a, b, c are distinct real numbers. Let
x3 - a
x4 - b
0
x5 + c
P(x) = x 3 + a
(b) a, b, c
(d) none of these
2
x n +1
0
8. Suppose a, b are two non-zero numbers. Let
D=
(b + 2016)2
12. Let n be an integer and x, y, z > 1. Suppose
y1 1
y2 1 ,
y3 1
then D2 is equal to
(a) 64
(c) 192
(a + 2016)2
a2
5. Let
(a) 0
(c) – 6
10. Suppose a, b, c ΠR. Let
a 2 + b2
a 3 + b3 ,
a 2 + b 2 a 3 + b3 a 4 + b 4
then D is equal to:
(a) 0
(b) ab
(d) a3b5 + a5b3
(c) a6 + b6
9. Suppose a, b, c > 1. Let
log a
log b
log c
D = log (2007a) log (2007 b) log (2007 c)
log (2017 a) log (20117 b) log (2017 c)
then D is equal to:
(a) 0
(b) log (4024 abc)
Ê 2017 ˆ
(d) none of these
(c) log Ë
2007 ¯
0
x4 + b x5 - c
A value of x satisfying P(x) = 0 is
(a) –(a + b + c)
(b) a + b + c
(c) a + b – c
(d) 0
14. Suppose a, b, c Œ R and abc π 0. Let
1+ a
1
1
D = 1 + b 1 + 2b
1
1 + c 1 + c 1 + 3c
1 1 1
+ + is equal to:
a b c
(a) 0
(b) –1
(c) –2
(d) –3
15. Suppose n and m are natural numbers such that
If D = 0, then
D=
xm
x m +2
x2m
1
xn
2n
x m + 5 x n +6 x 2 m + 5
Then a possible relationship between n and m is
(a) n = m + 2
(b) n = m + 1
(c) n = m
(d) n = m – 1
Determinants 4.33
LEVEL 1
Straight Objective Type Questions
Œ R and a + b + c π 0. Let
c+a a+b
a+b b+c
b+c c+a
16. Suppose a, b, c
b+c
D= c+a
a+b
If D = 0, then
(a) a = b = c
(c) a = b + c
3
3
3
(b) a + b – c = 0
(d) a = b = c = 0
p +2
+ 2a q + 2
n
2 +a
D=
2
2
n
2
p
2
2
q
D= b
e
yz
zx
zy
z2 + 1
a
D= x
x
and f(x) = (x
=5
(b) –1
(d) –3
x x
b x
x c
– a) (x – b) (x – c)
Determinant D is equal to:
(a) f(x) – x3
(c) xf ¢(x) – f(x)
(b) f ¢(x)
(d) f ¢(x) – x f ≤(x)
24. Straight line
n+2
n +1
2
+ 3b r + a
+b
n +1
2b
2
20. Suppose a, b, c, d, e and f are in G.P. with common
ratio > 1. Let p, q, r be three real numbers. Let
d2
y +1
23. Let
p +2 +a
q + 2 + 2b r - c
Then D equals:
(a) –1
(b) 0
(d) p2 q2 r2 – 4 (a + b + c)
(c) p2 q2 r2 – 3abc
a2
yx
(a) 0
(c) –2
19. Suppose a, b, c are in A.P. Let
2
xz
-1 cos C cos B
-1 cos A
D = cos C
cos B cos A
-1
then D equals:
3 (a + b + c) (bc + ca + ab)
a+b+c
3 (a + b + c) (a2 + b2 + c2)
0
n +1
xy
2
22. Suppose A, B, and C are angles of a triangle. Let
3a
-a + b -a + c
3b
-b + c
D = -b + a
-c + a -c + b
3c
Then D equals
2
x2 + 1
Then (x, y, z) lies on a
(a) plane
(b) straight line
(c) sphere
(d) none of these
17. Distance of line
x +1
x
x
x+2
x
y= x
x
x
x+3
from the origin is
6
7
(b)
(a)
11
13
6
7
(c)
(d)
122
122
18. Suppose a, b, c ΠR. Let
(a)
(b)
(c)
(d)
21. Suppose point (x, y, z) in space satisfies the equation
c2 f 2 r
Then D depends on
(a) a, b, c
(b) d, e, f
(c) p, q, r
(d) none of these
2-x-y
2x
2y
4
4
x-y-2
2x
=0
2y
y-2-x
passes through the fixed point
(a) (–2, –2)
(c) (0, –2)
(b) (–2, 0)
(d) ( –1, –1)
25. Suppose a ΠR. Let
x+a
x
x
f (x) =
x
x+a
x
x
x
x+a
Then f (2x) – f (x) is equal to
(a) 3xa2
(b) 3x2a
2
(d) a2x
(c) xa
26. If a, b, g are the roots of x3 + ax2 + b = 0, then
the determinant
4.34
Complete Mathematics—JEE Main
a b g
D = b g a equals
g a b
(a) – a3
(c) a2 – 3b
(b) a3 – 3b
(d) a3
27. If a, b, g are the roots of x3 + bx + c = 0, then the
determinant
a b g
D = b g a equals
g a b
(a) – b3
(c) b2 – 3c
28. If a, b, c are distinct
bc ca
D = ca ab
ab bc
(a)
(b)
(c)
(d)
1
cos x
D = -1 1 - cos x
(b) b3 – 3c
(d) 0
and different from zero and
ab
bc = 0, then
ca
a–1 + b–1 + c–1 = 0
a–1 + b–1 – c–1 = 0
a–1 – b–1 + c–1 = 0
a–1 – b–1 – c–1 = 0
l a l 2 + a2 1
29. The determinant D = l b l 2 + b2 1 equals
lc
l 2 + c2
sin 2 a
cos 2a
cos2 a
D = sin 2 b
cos 2b
cos2 b equals
sin 2 g
cos 2g
cos2 g
2
3
1 b
1 c2
b
1
b
1
c
a2
b2
c2
the value of l is
(a) – 1
(c) 9
(b) 0
(d) 18
equals
1 - 2 sin ( x + p / 4 )
(a) 0
(c) 1
(b)
(d)
x a
34. The factors of D = a x
a b
–1
none of these
b
b are
x
(a) x – a, x – b and x + a + b
(b) x – a, x – b and x – a – b
(c) x + a, x + b and x – a – b
(d) none of these
35. If q e R, maximum value of
1
1
1
1
is
D = 1 1 + sin q
1
1
1 + cos q
(a) 1/2
3 /2
(b)
(d) 3 2 /4
x+a
b
c
a
x+b
c
= 0, then x equals
36. If D =
a
b
x+c
(a) a + b + c
(b) – (a + b + c)
(c) 0, a + b + c
(d) 0, – (a + b + c)
37. The determinant
(c)
2
sec 2 q
tan q
12
1
=l a
c3
-1
2
(a) 0
(b) – 1
(c) sin2 a + sin2 b + sin2 g
(d) none of these
31. If bc + ca + ab = 18, and
a3
0
0
sin x + cos x
1
(a) l(a – b) (b – c) (c – a)
(b) l(a2 + b2 + c2)
(c) l(a + b + c)
(d) l2(a – b) (b – c) (c – a)
30. If a, b, g are real numbers, then the determinant
1 a2
32. If x π 0, the determinant
a0 a1 a2
D = -x x 0
0 -x x
vanishes if
(b) a0 + a1 = 2a2
(a) a0 + a1 + a2 = 0
(c) a0 + a2 = 2a1
(d) none of these
33. If x ΠR, the determinant
tan 2 q
2
sec q
10
1
-1 equals
2
2 sin2q
12 sec2 q – 10 tan2 q
12 sec2 q – 10 tan2 q + 5
0
- a 2b 0
38. If D = 0 - a 2b = 0, then
2b 0 - a
(a)
(b)
(c)
(d)
(a) 1/b is a cube root of unity
(b) a is one of the cube roots of unity
Determinants 4.35
(c) b is one of the cube roots of 8
(d) a/b is a cube root of 8
39. The determinant
1 1+ i
i
D = 1+ i
i
1 equals
i
1
6i -3i w
45. If D = 4 3i - w = x + iy, then
20 3 iw
(a) x = 3, y = 1
(c) x = 0, y = 3
p
p
£x£
46. If –
4
4
of D = 0
sin x
D = cos x
cos x
1+ i
(a) 7 + 4i
(b) – 7 + 4i
(c) – 7 – 4i
(d) 2(i – 1)
40. If a, b, c are non-zero real numbers, then
1 1
1 ab
+
a b
1 1
+
equals
D = 1 bc
b c
1 1
+
1 ca
c a
(a) 0
(c) a–1 + b–1 + c–1
(b) bc + ca + ab
(d) abc – 1
x + iy = -i
42. If D =
then l equals
(a) 0
(c) x
1
43. Let D = - sin q
-1
equals
(a)
(b)
(c)
(d)
-w
2
1
sin q , 0 £ q £ 2p.The
1
(b) D Œ (2, •)
(d) D Π[2, 4]
bc - ac
ab - a 2
a - b b2 - ab
bc - ac
c - a ab - a 2
(b – c) (c – a) (a – b)
abc (b – c) (c – a) (a – b)
(a + b + c) (b – c) (c – a) (a – b)
0
then
1
(a) x = – 1, y = 0
(c) x = 1, y = 1
(b) x = 1, y = – 1
(d) none of these
48. If eix = cosx + isinx and
(a) x = – 1, y =
b2 - ab b - c
44. The determinant D =
w
e -p i / 3
(b) 1
(d) 1 – x2
(a) D = 0
(c) D Π(2, 4)
w2
1
ep i / 4
x + iy = e-p i / 4
p + qx
a c p
px + q = l b d q
w
u v w
sin q
1
- sin q
cos x cos x
sin x cos x is
cos x sin x
1
0
loga b + logb c + logc a
logabc (a + b + c)
none of these
a + bx c + dx
ax + b cx + d
u
v
, the number of distinct real roots
(a) 0
(b) 2
(c) 1
(d) 3
47. If w π 1 is a complex cube root of unity, and
1
i
-w
log a ( abc ) log a b log a c
41. If a, b, c > 1, then D = logb ( abc )
logb c
1
logc ( abc ) logc b
1
equals
(a)
(b)
(c)
(d)
(b) x = 1, y = 3
(d) x = 0, y = 0
ep i / 3
e2p i / 3 , then
1
e-2p i / 3
e-2p i / 3
2
(b) x = 1, y = - 2
(c) x = - 2 , y = 2
(d) none of these
49. If a, b, c e R, the number of real roots of the
equation
x
c -b
-c x
a = 0, is
b -a x
(a) 0
(b) 1
(c) 2
(d) 3
1
x
x2
50. If D = x
x2
1
2
1
x
x
D1 =
= – 7 and
x3 - 1
0
0
x - x4
x - x4
x3 - 1
(a) D = 7
(c) D = – 49
x - x4
x 3 - 1 , then
0
(b) D = 343
(d) D = 49
4.36
Complete Mathematics—JEE Main
sin a
51. If D = sin b
sin g
cos a
cos a
cos a
sin a + cos b
sin b + cos b
sin g + cos b
57. For a fixed positive integer n, if
then D equals
(a) sin a sin b sin g
(b) cos a sec b tan g
(c) sin a sin b sin g + cos a cos b cos g
(d) 0
52. Suppose a, b, c ΠR and a, b, c > 0,
D=
2
sin 2 q
4 sin 4q
2
cos q
1 + sin q
4 sin 4q
cos2 q
sin 2 q
1 + 4 sin 4q
=0
are given by
(a) p/24, 5p/24
(b) 7p/24, 11p/24
(c) 5p/24, 7p/24
(d) 11p/24, p/24
54. The number of distinct roots of the equation
x2 - 1
x2 + 2 x + 1
P(x) = 2 x 2 + x - 1 2 x 2 + 5 x - 3
2 x2 + 3x + 1
4 x2 + 4 x - 3 = 0
6 x 2 - x - 2 6 x 2 - 7 x + 2 12 x 2 - 5 x - 2
is
(a) 6
(c) 3
(b) 5
(d) 4
2rp
2rp
+ i sin
, then value of the
55. If ar = cos
9
9
determinant
1 a8 a7
D = a3 a2 a1 is
a6
a5
a4
(a) – 1
(b) 1
(c) 0
(d) – 2
56. If a π p, b π q, c π r and the system of equations
px + by + cz = 0
ax + qy + cz = 0
ax + by + rz = 0
has a non-zero solution, then value of
p+a q+b r+c
+
+
is
p-a q-b r-c
(a) 2
(c) 1
(b) – 2
(d) 1
D
is equal to
n! ( n + 1)! ( n + 2 )!
(a) – 4
(b) – 2
(c) 2
(d) 4
x+2 x+7 a
58. If a, b, c are in A.P., and D = x + 5 x + 11 b
x + 8 x + 15 c
then D equals
(a) 0
(b) 1
(c) – (a + b + c)
(d) a + b + c.
then
log a
log b
log c
Let D = log (7 a) log (49 b) log (343 c)
log (3 a ) log (9 b) log (27 c)
then D is equals to
(a) 0
(b) –1
(c) 1
(d) 30
53. The values of q lying between q = 0 and q = p/2
and satisfying the equation
1 + cos2 q
n!
(n + 1)! (n + 2)!
D = ( n + 1)! ( n + 2 )! ( n + 3)!
(n + 2)! (n + 3)! (n + 4)!
1+ y 1- y 1- y
59. If D = 1 - y 1 + y 1 - y = 0, then value of y are
1- y 1- y 1+ y
(a) 0, 3
(c) – 1, 3
60. The determinant
(b) 2, – 1
(d) 0, 2
al + a ¢l ¢ am + a ¢m ¢ an + a ¢n ¢
D = bl + b ¢l ¢ bm + b ¢m ¢ bn + b ¢n ¢
cl + c ¢l ¢ cm + c ¢m ¢ cn + c ¢n ¢
is equal to
(a) (abc + a¢b¢c¢) (lmn + l¢m¢n¢)
(b) abc lmn + a¢b¢c¢ l¢m¢n¢
(c) (a2 + b2 + c2) (l2 + m2 + n2)
+ (a¢2 + b¢2 + c¢2) (l¢2 + m¢2 + n¢2)
(d) 0
61. If a = i, b = w, c = w2 where w is complex cube
root of unity, then
a
a+b
a+b+c
D = 3a 4 a + 3b 5a + 4b + 3c is equal to
6 a 9a + 6b 11a + 9b + 6c
(a) – w
(c) i
1
62. If D =
m
C1
1
m +3
C1
(b) – w2
(d) – i
1
m +6
C1
= 2a 3b 5g, then
C2 m + 3 C2 m + 6 C2
a + b + g is equal
(a) 3
(b) 5
(c) 7
(d) none of these
63. Suppose a, b, c, x, y ΠR. Let
1 2 + ax 3 + ay
D = 1 2 + bx 3 + by
1 2 + cx 3 + cy
m
Determinants 4.37
Then D is independent of
(a) a, b, c
(b) x, y
(c) a, b, c, y
(d) a, b, c, x, y
64. If A, B and C are the angles of a triangle, then the
determinant
0
cos C cos B
-1 cos A is equal to
D = cos C
cos B cos A
-1
(a) sin2A
(c) sin2C
65. Let f(x) =
given by
cos x
(b) sin2B
(d) 0
x 1
2 sin x
tan x
x2
x
f (x)
is
2 x , then lim
2
xÆ0 x
1
(a) 0
(b) – 1
(c) 2
(d) 3
66. If w is a complex cube root of unity, then value of
a1 + b1w
D = a2 + b2w
a1w 2 + b1
c1 + b1w
2
c2 + b2w
a2w + b2
2
a3 + b3w a3w + b3 c3 + b3w
is
(a) 0
(b) – 1
(c) 2
(d) none of these
67. If pqr π 0 and the system of equations
(p + a)x + by + cz = 0
ax + (q + b)y + cz = 0
ax + by + (r + c)z = 0
a b c
has a non-trivial solution, then value of + +
p q r
is
(a) – 1
(b) 0
(c) 1
(d) 2
68. The system of equations
ax + by + (aa + b)z = 0
bx + cy + (ba + c)z = 0
(aa + b)x + (ba + c)y = 0
has a non-zero solutions if a, b, c are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.
69. If the system of equations
ax + ay – z = 0
bx – y + bz = 0
– x + cy + cz = 0
(where a, b, c π – 1) has a non-trivial solution, then
1
1
1
is
value of
+
+
1+ a 1+ b 1+ c
(a) 2
(b) – 1
(c) – 2
(d) 0
70. The values of l for which the system of equations
(l + 5)x + (l – 4)y + z = 0
(l – 2)x + (l + 3)y + z = 0
lx + ly + z = 0
has a non-trivial solution is (are)
(a) – 1, 2
(b) 0, – 1
(c) 0
(d) none of these
71. Given the system of equations
(b + c) (y + z) – ax = b – c
(c + a) (z + x) – by = c – a
(a + b) (x + y) – cz = a – b
(where a + b + c π 0); then x : y : z is given by
(a) b – c : c – a : a – b (b) b + c : c + a : a + b
a b c
(c) a : b : c
(d) : :
b c a
72. If a, b, c e R and a + b + c π 0 and the system of
equations
ax + by + cz = 0
bx + cy + az = 0
cx + ay + bz = 0
has a non-zero solution, then a : b : c is given by
(a) 1 : a : b
where a, b are roots of ax2 + bx + c = 0
(b) 1 : r : r2 where r is some positive real number
(c) 1 : k : 2k where k is some positive real number
(d) none of these
73. If p is a constant and
x3
f(x) = 1
p
x3 3x 2
-6 4 , then f ¢≤(x) is
p2
p3
(a) proportional to x3 (b) proportional to x2
(c) proportional to x (d) a constant
a1
74. If D = a2
a3
b1
b2
b3
a1 + pb1
D = a2 + pb2
a3 + pb3
c1
c2
c3
and
b1 + qc1
b2 + qc2
b3 + qc3
c1 + ra1
c2 + ra2
c3 + ra3
then
(a) D = D
(b) D = D (1 – pqr)
(c) D = D(1 + pqr)
(d) D = D(1 + p + q + r)
75. Number of real values of l for which the system
of equations
(l + 3)x + (l + 2)y + z = 0
3x + (l + 3)y + z = 0
2x + 3y + z = 0
has a non-trivial solutions is
(a) 0
(b) 1
(c) 2
(d) infinite
4.38
Complete Mathematics—JEE Main
Assertion-Reason Type Questions
76. Statement-1: If
D(x) =
ax + by + cz = 0
sin x - cos x
pˆ
Ê
sin Á x - ˜
Ë
3¯
pˆ
Ê
cos Á x - ˜
Ë
3¯
Êp
ˆ
sin Á - x˜
Ë3
¯
pˆ
Ê
tan Á x - ˜
Ë
4¯
pˆ
Ê
sec Á x - ˜
Ë
4¯
Ê 2p
ˆ
cos Á
+ x˜
Ë 3
¯
Ê 2p
ˆ
sec Á
+ x˜
Ë 3
¯
pˆ
Ê
cot Á x + ˜
Ë
4¯
Êpˆ
then D Á ˜ = 0
Ë 4¯
Statement-2: If A is a skew symmetric matrix of
odd order, then |A| = 0.
77. Statement-1: Let p < 0 and a1, a2, . . . a9 be the
nine roots of x9 = p, then
a1
D = a4
a7
a2 a3
a5 a6 = 0
a8 a 9
Statement-2: If two rows of a determinant are
identical, then determinant equals zero.
78. Statement-1:
6
w =
2i
12
3 + 8i
18
2 + 12i
3 + 6i
3 2 + 6i
27 + 2i
is a purely imaginary number.
Statement-2: |z| = | z | for each complex number z.
79. Statement-1: If ai, bi ΠN, for i = 1, 2, 3 and
(1 + x )a1 b1
(1 + x )a1 b2
(1 + x )a1 b3
a b
D(x) = (1 + x ) 2 1
(1 + x )a2 b2
(1 + x )a2 b3 ,
(1 + x )a3 b1
(1 + x )a3 b2
(1 + x )a3 b3
then coefficient of x in expansion of D(x) is 0.
Statement-2: If P(x) = (1 + x)n, n Œ N then coefficient of x in the expansion of P(x) is P¢(0).
80. Statement-1: If a + b + c = 0 and a2 + b2 + c2
π bc + ca + ab, then the system of homogenous
equations
bx + cy + az = 0
cx + ay + bz = 0
has infinite number of solutions.
Statement-2: If |A| = 0, the system of equations
AX = B has infinite number of solutions.
81. Suppose a, b, c ΠR. Consider the system of linear
equations:
ax + by + cz = 0
bx + cy + az = 0
cx + ay + bz = 0
Statement-1: If a + b + c π 0 and a2 + b2 + c2
= bc + ca + ab, then the system of equations has
infinite number of solutions.
Statement-2: If a + b + c = 0 and a2 + b2 + c2
π bc + ca + ab, then the system of equations has
unique solutions.
82. Suppose a1, a2, a3, b1, b2, b3 ΠR.
Let
a1 b1 1
D1 = a2 b2 1
a3 b3 1
2 a1b1
D2 = a1b2 + a2 b1
a1b3 + a3 b1
a1b2 + a2 b1
2 a2 b2
a2 b3 + a3 b2
a1b3 + a3 b1
a2 b3 + a3 b2
2a3 b3
Statement-1: D2 π 0 for some values of a1, a2, a3,
b1, b2, b3 ΠR.
Statement-2: If (a1, b1), (a2, b2), (a3, b3) are noncollinear than D1 π 0.
83. Suppose a1, b1, c1, d1 a2, b2, c2, d2 are eight integers.
Statement-1: ( a12 + b12 + c12 + d12 ) ( a22 + b22 + c22 + d22 )
can be written in the form a2 + b2 + c2 + d2 where
a, b, c, d are some integers.
Statement-2: If a, b, c, d ΠZ, then
a + ib c + id
a2 + b2 + c2 + d2 =
-c + id a - ib
Determinants 4.39
LEVEL 2
Straight Objective Type Questions
84. If l12 + m12 + n12 = 1 etc., and l1l2 + m1m2 + n1n2
= 0 etc., and
l1 m1 n1
D = l2 m2 n2
l3 m3 n3
then
(a) | D | = 3
(b) | D | = 2
(c) | D | = 1
(d) D = 0
85. If a, b, c are non-zero real number and
b2 c 2
bc
b+c
D = c2 a2
ca
c+a
2 2
a b
then D equals
(a) abc
(c) bc + ca + ab
ab a + b
86. Let D(x) =
2
3x + x + 9
2x - 5
3
6x + 1
9
7 x 2 - 6 x + 9 14 x - 6 21
= ax3 + bx2 + cx + d,
then a equals
(a) – 1
(b) 0
(c) 2
(d) none of these
87. The number of distinct values of t for which the
system
(a + t)x + by + cz = 0
ax + (b + t)y + cz = 0
ax + by + (c + t)z = 0
has a non-trivial solution is
(a) 1
(b) 2
(c) 3
(d) none of these
88. If a2 + b2 + c2 = 1, then
a 2 + (b2 + c 2 ) cos q
ab (1 - cos q )
2
ba (1 - cos q )
ca(1 - cos q )
2
2
b + (c + a ) cos q
cb(1 - cos q )
ac (1 - cos q )
bc (1 - cos q )
2
2
2
c + (a + b )cos q
(a) cos2q
(c) 1
(b) 0
(d) sin2q
A(a, b, g, q) =
cos (b + q ) sin (b + q ) 1
cos (g + q ) sin (g + q ) 1
p
p 2p
Numerical value of A Ê - , 0, , ˆ is
Ë 2
2 13 ¯
(a) 0
(b) – 1
(c) 2
(d) none of these
90. If a, b, c are positive integers such that a > b > c
and
1 1 1
a b c =–2
a 2 b2 c 2
then 3a + 7b – 10c equals
(a) 10
(b) 11
(c) 12
(d) 13
91. If A, B, C, P, Q, R ΠR, and
(b) a2 b2 c2
(d) 0
x2 - 5x + 3
89. Suppose a, b, g, q ΠR and
cos (a + q ) sin (a + q ) 1
equals
cos ( A + P ) cos ( A + Q ) cos ( A + R)
D = cos ( B + P ) cos ( B + Q) cos ( B + R)
cos (C + P ) cos (C + Q) cos (C + R)
(a) D depends on P, Q, R
(b) D depends on A, B, C
(c) D depends on A, B, C, P, Q, R
(d) none of these
2 cos x
1
0
then
92. Let f (x) =
1
2 cos x
1
0
1
2 cos x
(a) f Ê p ˆ = 1
(b) f ¢ Ê p ˆ = - 3
p
(c) f Ê ˆ = – 1
Ë2 ¯
(d) none of these
Ë 3¯
Ë 3¯
93. Consider the set A consisting of all determinants
of order 3 with entries 0 and 1 only. Let B be the
subset of A consisting of all the determinants with
value 1. Let C be the subset of A consisting of all
the determinants with value –1. Then
(a) C = f
(b) B has as many elements as C
(c) A = B « C
(d) A = B » C
94. let w = e2pi/3 and consider the system of linear
equations
4.40
Complete Mathematics—JEE Main
x+y+z=a
x + w y +w2 z = b
x + w2 y + w z = c
If x, y, z is a solution of the above system of equa| a |2 + | b |2 + | c |2
is
tions, then value of
| x |2 + | y |2 + | z |2
(a) 9
(b) 6
(c) 3
(d) 1
a a 2 a3 - 1
b b2
b3 - 1
c c2
(b) 1
(d) –2
c3 - 1
95. If a, b, c are distinct and
then abc equals
(a) 0
(c) –1
=0
96. If the adjoint
Ê1
Á2
Á
Ë1
of
4
1
1
a 3 ¥ 3 matrix P is
4ˆ
7˜ then
˜
3¯
the possible value(s) of determinant P is (are)
(a) ± 2
(b) ± 3
(d) ± 1
(d) 0
1
-3
4
= 0, then x equals
97. If -5 x + 2
2
4
1
x-6
(a) 17, 21
(b) 0,19
(c) 0, 35
(d) 21, 35
Previous Years' AIEEE/JEE Main Questions
1. If the equation ax2 + 2bx + c = 0 has equal roots
then the determinant
D=
a
b
ax + b
b
c
bx + c
ax + b bx + c
0
is
(a) positive
(c) 0
(b) negative
(d) dependent on a.
[2002]
2. If l, m, n > 0 and l, m, n are the pth, qth, rth terms
of a G.P., then determinant
log l p 1
[2002]
D = log m q 1 equals
log n
w
1
log an +1
log an + 2
log an + 1
log an + 2
log an + 3
log an + 2
log an + 3
log an + 4
1 + a2 x
f(x) = (1 + a 2 ) x
1
equals
wn
(a) 1
(b) 2
(d) 0
[2003]
(c) w2
4. If the system of linear equations
x + 2ay + az = 0, x + 3by + bz = 0, x + 4cy + cz
= 0 has a non-zero solution, then a, b, c
(a) are in G.P
(b) are in H.P.
w
2n
2n
log an
is
(a) 2
(b) 1
(c) 0
(d) – 2
6. If a2 + b2 + c2 = – 2 and
r 1
(a) 0
(b) – 1
(c) p + q + r
(d) none of these
3. If 1, w, w2 are the cube roots of unity then
1
w n w 2n
D = wn
(c) satisfy a + 2b + 3c = 0
(d) are in A.P.
[2003]
5. If a1, a2, a3, ... , an, ... are in G.P., then the value
of the determinant
[2004, 2005]
(1 + b2 ) x (1 + c 2 ) x
1 + b2 x
(1 + c 2 ) x
(1 + a 2 ) x (1 + b2 ) x
then f(x) is a polynomial of degree
(a) 3
(b) 2
(1) 1
(d) 0
7. The system of equations
ax + y + z = a – 1
1 + c2 x
[2005]
x + ay + z = a – 1
has
(a)
(b)
(c)
(d)
x + y + az = a – 1
no solution, if a is
not – 2
1
–2
either – 2 or 1
[2005]
Determinants 4.41
1
1
1
8. If D = 1 1 + x
1 for x π 0, y π 0, then D is
1
1
1+ y
(a)
(b)
(c)
(d)
9. Let
divisible by neither x nor y
divisible by both x and y
divisible by x but not y
divisible by y but not x
a, b, c be such that b(a + c) π 0. If
a a +1 a -1
D = -b b +1 b -1
c c -1 c +1
+
a +1
a -1
b +1
b -1
n+ 2
n+1
(-1)
a (-1)
[2007]
c -1
c +1
=0
n
b (-1) c
[2009]
of a scalene triangle, then the
c
a is:
b
[2014]
13. If a, b, c are non-zero real numbers and if the
system of equations
(a – 1)x = y + z,
(b – 1)y = z + x,
(c – 1)z = x + y,
(a) a + b + c
(b) abc
(c) 1
(d) –1
[2014 online]
14. Let for i = 1, 2, 3, pi(x) be a polynomial of degree
2 in x, pi¢(x) and pi≤(x) be the first and second order
derivatives of pi(x) respectively. Let,
p1¢ ( x )
p2¢ ( x )
p3¢ ( x )
È p1 ( x )
Í
A(x) = p2 ( x )
Í
ÍÎ p3 ( x )
p1¢¢( x ) ˘
p2¢¢( x )˙
˙
p3¢¢( x ) ˙˚
and B(x) = [A(x)]T A(x). Then determinant of B(x).
(a)
(b)
(c)
(d)
is a polynomial of degree 6 in x
is a polynomial of degree 3 in x
is a polynomial of degree 2 in x
does not depend on x
[2014 online]
a2
has a non-trivial solution for only one value of a
lying in the interval (0, p /2).
Statement-2: The equation in a
sin a
cos a
- sin a
(d) ab
15. If
[2013 online]
11. Statement-1: The system of linear equations
x + (sin a)y + (cos a)z = 0
x + (cos a)y + (sin a)z = 0
x – (sin a)y – (cos a)z = 0
cos a
D = sin a
cos a
(b) 1
has a non-trivial solution, then ab + bc + ca equals:
then the value of n is
(a) any odd integer
(b) any integer
(c) zero
(d) any even integer
10. If a, b, c are sides
a b
value of D = b c
c a
(a) non-negative
(b) negative
(c) positive
(d) non-positive
1
ab
(c) –1
(a)
cos a
sin a = 0
- cos a
has only one solution lyining in the interval
(0, p /2).
[2013 online]
n
n
12. If a, b π 0, and f(n) = a + b and
3
1 + f (1) 1 + f (2)
1 + f (1) 1 + f (2) 1 + f (3)
1 + f (2) 1 + f (3) 1 + f (4)
= K(1 – a)2 (1– b)2 (a – b)2, then K is equal to:
b2
c2
a 2 b2 c 2
(a + l )2 (b + l )2 (c + l )2 = kl a b c , l π 0,
1 1 1
( a - l )2 ( b - l )2 ( c - l )2
then k is equal to:
(a) 4labc
(c) 4l2
(b) – 4labc
(d) – 4l2
[2014 online]
16. If Dr =
r
n
2
2r - 1
3r - 2
n -1
a
1
n(n - 1) (n - 1)2
2
1
(n - 1)(3n + 4)
2
n -1
then the value of
 Dr :
r =1
(a)
(b)
(c)
(d)
depends only on a
depends only on n
depends both on a and n
is independent of both a and n.
[2014 online]
Complete Mathematics—JEE Main
4.42
17. The set of all values of l for which the system of
linear equations:
2x1 – 2x2 + x3 = lx1
2x1 – 3x2 + 2x3 = lx2
– x1 + 2x2 = lx3
has a non-trivial solution,
(a) is an empty set
(b) is a singleton
(c) contains two elements
(d) contains more than two elements
x2 + x
18. If
x +1
2
D = 2 x + 3x - 1
3x
2
x + 2x + 3
then a is equal to:
(a) 12
(c) – 12
*
19 . The least value of
x
determinant D = 1
1
[2015]
x-2
3 x - 3 = ax - 12,
2x - 1 2x - 1
(b) 24
(d) – 24
the
1
y
1
(a) -2 2
(b) -16 2
(c) – 8
(d) – 1
[2015 online]
20. The system of linear equations
x + ly – z = 0
lx – y – z = 0
x + y – lz = 0
has a non-trivial solution for
l
(b) exactly one value of l
(c) exactly two values of l
(d) exactly three values of l
[2016]
21. The number of distinct real roots of the equation.
cos x sin x sin x
D = sin x cos x sin x = 0
sin x sin x cos x
È p p˘
in the interval Í - , ˙ is:
Î 4 4˚
(a) 1
(b) 4
(c) 2
(d) 3
[2015 online]
product xyz for which the
1
1 is non-negative, is:
[2016 online]
z
Previous Years’ B-Architecture Entrance
Examination Questions
1. If the system of equations
x + y + z= 0
ax + by + z = 0
bx + y + z = 0
has a non-trivial solution, then
(a) b2 = 2b + 1
(c) b – a = 0
(b) b2 = 2b – 1
(d) b2 = 2b
n
then the value of
k =1
(a)
(b)
(c)
(d)
[2006]
only on a and b not on g
on all a, b and g
on none of a, b and g
only on a, not on b and g
[2007]
4. Let A = (aij) be a 3 × 3 matrix whose determinant
is 5. The determinant of the matrix B = (2i – j aij) is
(a) 5
(b) 10
(c) 20
(d) 40
[2008]
r -1
r -1
r -1
2
2(3 ) 4(5 )
, for
b
g
5. Let Dr = a
2. The system of linear equations
(l + 3)x +(l + 2)y + z = 0
3x + (l + 3)y + z = 0
2x + 3y + z = 0
has a non-trivial solution
(a) if l = 1
(b) if l = –1
(c) for no real value of l
(d) if l = 0
 D k depends
2n - 1
3n - 1
5n - 1
n
r = 1, 2,…, n. Then  D r is
[2007]
r =1
3. If
(a)
(b)
(c)
(d)
2(3k -1 ) 3(4 k -1 ) 4(5k -1 )
Dk =
a
b
g
3n - 1
4n - 1
5n - 1
*
independent of a, b, g and n
independent of n only
depends on a, b, g and n
independent of a, b, g only
Question is incorrect as xyz can take any real value.
[2010]
Determinants 4.43
6. If x1, x2, x3,…, x13 are in A.P. then the value of
e x1
e x4
e x7
e x4
e x7
e x10 is:
Concept-based
e x7 e x10 e x13
(a) 27
(b) 0
(c) 1
(d) 9
7. The value of the determinant
[2011]
2 5
15 + 26
5
10 is equal to:
3 + 65
15
5
5
(b) 5 3 ( 6 - 5 )
(c) 5 ( 6 - 5)
(d) 3 ( 6 - 5 ) [2012]
8. If the system of linear equations, x + 2ay + az =
0, x + 3by + bz = 0 and x + 4cy + cz = 0 has a
non-zero solution, then a, b, c satisfy:
(a) 2b = a + c
(b) b2 = ac
(c) 2ac = ab + bc
(d) 2ab = ac + bc [2013]
9. In a DABC, if
1 a b
1 c a = 0,
1 b c
then sin2A + sin2B + sin2C is
3
9
3
(a)
(b)
2
4
5
(c)
(d) 2
[2014]
4
10. The system of linear equations
x – y + z= 1
x + y – z= 3
x – 4y + 4z = a has:
(a) a unique solution when a = 2
(b) a unique solution when a π – 2
(c) an infinite number of solutions, when a = 2
(d) an infinite number of solutions, when a = –2
[2015]
Ê pˆ
11. For all values of q Œ Á 0, ˜ , the determinant of the
Ë 2¯
È -2
Í
matrix D = Í- sin q
Í -3
Î
tan q + sec2 q
cos q
-4
lies in the interval:
(a) [3, 5]
(b) (4, 6)
Ê 5 19 ˆ
(c) Á , ˜
Ë2 4 ¯
È 7 21 ˘
(d) Í , ˙
Î2 4 ˚
1.
5.
9.
13.
(d)
(a)
(a)
(d)
2.
6.
10.
14.
(d)
(c)
(b)
(d)
3.
7.
11.
15.
(a)
(d)
(d)
(a)
4. (d)
8. (a)
12. (c)
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
57.
61.
65.
69.
73.
77.
81.
(c)
(c)
(a)
(a)
(c)
(d)
(a)
(d)
(b)
(b)
(c)
(d)
(b)
(a)
(d)
(d)
(c)
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
62.
66.
70.
74.
78.
82.
(a)
(a)
(d)
(a)
(a)
(d)
(d)
(c)
(d)
(d)
(a)
(a)
(a)
(d)
(c)
(b)
(d)
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
63.
67.
71.
75.
79.
83.
85.
89.
93.
97.
(d)
(c)
(b)
(c)
86. (b)
90. (d)
94. (c)
Level 1
13 + 3
(a) 5 3 ( 6 - 5)
Answers
3 ˘
˙
sin q ˙ always
3 ˙˚
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
64.
68.
72.
76.
80.
(a)
(d)
(d)
(a)
(a)
(d)
(a)
(d)
(d)
(a)
(d)
(d)
(a)
(b)
(d)
(a)
(c)
(b)
(c)
(d)
(d)
(a)
(d)
(d)
(a)
(d)
(c)
(a)
(d)
(a)
(a)
(a)
(a)
(a)
Level 2
84.
88.
92.
96.
(c)
(a)
(b)
(a)
87. (b)
91. (d)
95. (b)
Previous Years’ AIEEE/JEE Main Questions
1.
5.
9.
13.
17.
21.
(c)
(c)
(a)
(b)
(c)
(c)
2.
6.
10.
14.
18.
(a)
(b)
(b)
(d)
(b)
3.
7.
11.
15.
19.
(d)
(c)
(b)
(c)
(?)
4.
8.
12.
16.
20.
(b)
(b)
(b)
(d)
(d)
Previous Years’ B-Architecture Entrance
Examination Questions
1. (b)
5. (a)
9. (b)
2. (c)
6. (b)
10. (d)
3. (c)
7. (c)
11. (a)
4. (a)
8. (c)
Hints and Solutions
Concept-based
1. det (adj (A)) A–1) = det (adj(A)) det(A–1)
[2016]
= (det(A))2
1
= det(A)
det( A)
4.44
Complete Mathematics—JEE Main
2. Use R1 Æ R1 + R3 – 2R2
2
3. Using R2 Æ R2 – x R1 and R3 Æ R3 – xR1, we get
1
x
a2
D = 0 1 - x3
0
x(1 - x 3 )
1 - x3
0
= (1 – x3)2 = (1 – x)2 (1 + x + x2)2
As x π 1, D = 0 fi 1 + x + x2 = 0 fi x = w, w2
4. Taking 10!, 11! and 12! common from C1, C2, C3
respectively, we get
D = (10! 11! 12!) D1
where
1
1
1
D1 =
11
12
13
(11)(12) (12)(13) (13)(14)
Using C3 Æ C3 – C2 and C2 Æ C2 – C1, we get
1
0
0
D1 =
11
1
1 =2
(11)(12) 2(12) 2(13)
Thus,
D
(10!)3
= 2(11) (12) = 264
D2 = 192
fi
8. Write
1+1
a+b
a 2 + b2
a+b
a 2 + b2
a 3 + b3
a 2 + b2
a 3 + b3
a 4 + b4
1
= a
1
b
0
0
1
a
1
b
0
0 =0
a2
b2
0
a2
b2
0
9. Use R2 Æ R2 – R1, R3 Æ R3 – R1 and
log x – log y = log (x/y) to obtain
log a
log b
log c
D = log (2007) log (2007) log (2007) = 0
log (2017) log (2017) log (2017)
[
R2 and R3 are proportional]
10. Using R2 Æ R2 – R1, we get
Applying R1 Æ R1 – 2 (2016) R2 – R3, we get
1 1 1
3
D = - 4 (2016) a b c
a2
b2
c2
= – 4 (2016)3 (a – b) (b – c) (c – a)
11. Using R2 Æ R2 + R1, R3 Æ R3 – R1, we get
x
-6
-1
P(x) = x + 2 -3( x + 2) x + 2
-3 - x 2( x + 3) x + 3
x
= (x + 2) (x + 3) 1
-1
Using R1 Æ R1 + R3, R2 Æ
-6
-3
2
R2 –
-1
1
1
R3, we get
x -1 -4 0
-5 0
P(x) = (x + 2) (x + 3) x
-1
2 1
12. Write D as
1 x
x2
D = xn+1 yn+1 zn+1 1 y
y2
1 z z2
= (xyz)n+1 (x – y) (y – z)(z – x)
7. Use R1 Æ R1 + R3 – 2R2.
D=
c2
sum of zeros of P(x) is –2 – 3 + 13/5 = –12/5.
1
3 2
( 4) = 4 3
6.
D = area of triangle =
2
4
|D| = 8 3
b2
= (x + 2) (x + 3) (– 5x + 13)
5. Product of zeros = –P(0) = 0
fi
(a + 2016)2 (b + 2016)2 (c + 2016)2
a
b
c
D = - 4(2016)
\
n+1 = 2
fi n = 1.
13. When x = 0, P(0) is a skew-symmetric determinant
of odd order.
14. Write D = D1 + aD2 where
1
1
1
D1 = 1 + b 1 + 2 b
1
1 + c 1 + c 1 + 3c
1
0
0
1
and D2 = 1 + b 1 + 2b
1 + c 1 + c 1 + 3c
= (1 + 2b) (1 + 3c) – (1 + c)
= 2 (b + c + 3c)
In D1, apply C2 Æ C2 – C1 and C3 Æ C3 – C1 to
obtain
1
0 0
D1 = 1 + b b -b = 2bc
1 + c 0 2c
Determinants 4.45
Thus, D = 2(bc + ac + ab + 3abc)
1 1 1
\
D =0 fi
+ + = –3
a b c
m
15. Taking x common from R1 and x
get
1 x2
xm
D = x 2m +5 1
xn
n-m
19. Using R1 Æ R1 – R3 – R2 we get R1 consists of all
zeros.
m+5
from R3, we
2n
20. If R is the common ratio of the G.P., then b = aR,
c = aR2, d = aR3, e = aR4, f = aR5, and
1
2
2
6
D = (a )(a R ) R
Level 1
16. Using C1 Æ C1 + C2 + C3, we get
1 c+a a+b
D = 2(a + b + c) 1 a + b b + c
1 b+c c+a
Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get
1 c+a a+b
D = 2(a + b + c) 0 b - c c - a
0 b-a c-b
2
= – 2(a + b + c) [(b – c) + (b – a) (c – a)]
= – (a + b + c) [(b – c)2 + (c – a)2 + (a – b)2]
Now, use D = 0, a + b + c π 0.
17. Write y = D1 + D2 where
1
x
x
x = 5x + 6
D1 = 0 x + 2
0
x
x+3
1
x
x
1 0 0
x = x 1 2 0 = 6x
and D2 = x 1 x + 2
1
x
x+3
1 0 3
Thus, y = 11x + 6
6
Its distance from the origin =
122
18. Using C1 Æ C1 + C2 + C3, we get
1 -a + b -a + c
3b
-b + c
D = (a + b + c) 1
1 -c + b
3c
Now, use R2 Æ R2 – R1, R3 Æ R3 – R1, we get
1 -a + b -a + c
D = ( a + b + c ) 0 2b + a - b + a
0 - c + a a + 2c
D = (a+b+c) [(a + 2b) (a + 2c) – (a – c) (a – b)]
= 3(a + b + c) (bc + ca + ab)
R
R4
[
m
1 x
x
If n – m = 2, then R1 and R3 are identical and hence
D = 0.
1
2
p
2
q =0
R4 r
C1 and C2 are identical]
21. Multiply C1 by x, C2 by y and C3 by z to obtain
5=
( x 2 + 1) x
xy 2
xz 2
yx 2
( y 2 + 1) y
yz 2
1
xyz
zx 2
zy 2
z( z 2 + 1)
Taking x common from R1, y from R2 and z from
R3, we get
x2 + 1
5=
xyz
xyz
x2
y2
z2
y2 + 1 z 2
x2
y2
z2
Using C1 Æ C1 + C2 + C3, we get
5 = (x2 + y2 + z2 + 1) D1
1
y2
where D1 = 1 y 2 + 1
1
y2
z2
z2
z2 + 1
Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get
1 y2
D1 = 0 1
0 0
z2
0 =1
1
Thus, x2 + y2 + z2 = 4, which represents a sphere.
22. Apply R1 Æ aR1 + bR2 + cR3 and use a = b cos C
+ c cos B etc.
23. Write D = D1 + xD2 where
a-x x x
D1 = 0
b x = (a - x )(bc - x 2 )
0
x c
1 x x
and D2 = 1 b x
1 x c
1
x
x
0
= 0 b-x
0
0
c-x
= (b – x) (c – x)
[use R2 Æ R2 – R1,
R 3 Æ R 3 – R 1]
4.46
Complete Mathematics—JEE Main
Thus, D = (a – x) (bc – x2) + x(b – x) (c – x)
= 2x3 – (a + b + c)x2 + abc
Also, x f ¢(x) – f(x)
= x [(x – b) (x – c) + (x – a) (x – c)
+ (x – a) (x – b)] – f (x)
= 2x3 – (a + b + c)x2 + abc
24. Applying R1 Æ R1 + R2 + R3, we get
1
0 = ( x + y + 2) 2 x
2y
1
1
x-y-2
2x
2y
y-2- x
1
0
0
0
0 = ( x + y + 2) 2 x - ( x + y + 2 )
2y
0
- ( x + y + 2)
0 = (x + y + 2)3
D = l b b2 1
c2 1
Using R3 Æ R3 – R2, R2 Æ R2 – R1, we get
c
25. Write f(x) = D1 + xD2 where
x
x+a
x
2
x
x = a[(x + a)2 – x2]
x+a
a2
a
c-b
x
x
x+a
2
0
2
2
0
c -b
1 b+a
1 c+b
= l(b – a) (c – b) (c – a)
= l(b – a) (c – b)
= l(a – b) (b – c) (c – a)
30. Use C2 Æ C2 – C3 + C1 to show that C2 consists
of all zeros.
31. We know
1 1 1
a b c = (a – b) (b – c) (c – a)
1 x x
= 0 a 0 = a2
0 0 a
f (x) = 2xa2 + a3 + xa2
= 3xa2 + a3
a2
\ f(2x) – f(x) = 3xa2
26. a + b + g = – a, b g + g a + a b = 0, a b g = – b.
Applying C1 Æ C1 + C2 + C3, we get
1 b g
D = (a + b + g ) 1 g a
1 a b
(1)
Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get
1
b
D= – a 0 g -b
0 a -g
1
2
D= l b-a b -a
3
= 2 xa + a
1
x
and D2 = 1 x + a
1
x
Thus,
a=
a a2 1
fi
x+y+2=0
It passes through (–1, –1).
a
D1 = 0
0
1
1
1
,b= ,g=
a
b
c
From (2) in solution question 26, we have
D = (abc)2 (a + b + g) [(b – g)2 + (g – a)2
+ (a – b)2]
fi a + b + g = 0 fi a–1 + b–1 + c–1 = 0
29. Using C2 Æ C2 – l2C3, we can write
where
Using C2 Æ C2 – C1, C3 Æ C3 – C1, we get
fi
= – a[b g + g a + a b – (a 2 + b 2 + g 2)] (2)
= – a[0 – a2] = a3
27. In this case a + b + g = 0.
Thus, from (1) in solution to question 26, we get
D = 0.
28. We can write
a b g
2
D = (abc) b g a
g a b
g
a -g
b -a
= – a[(g – b ) (b – a) – (a – g )2]
b2
c2
1 a2
a3
Let D1 = 1 b2
b3
1 c2
c3
Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get
1
a2
a3
D1 = 0 b 2 - a 2
b3 - a 3
0 c 2 - b2
c 3 - b3
= (b – a) (c – b)
b + a b2 + a 2 + ba
c+b
c 2 + b2 + cb
Determinants 4.47
3
Thus, D = 0 fi Ê a ˆ = 8
Applying C2 Æ C2 – bC1, we get
D1 = (b – a) (c – b)
= (b – a) (c – b)
Ë b¯
b + a a2
c+b
c
2
a2
b+a
c - a c2 - a2
[using R2 Æ R2 – R1]
b + a a2
c+a
1
= (a – b) (b – c) (c – a) (bc + ca + ab)
Thus,
l = 18
32. Show that D = (a0 + a1 + a2)x2
33. Using R2 Æ R2 + R1, we get
= (b – a) (c – b) (c – a)
1 cos x
0
D= 0
1
sin + cos x
0 - 1 1 - (sin x + cos x )
Using R3 Æ R3 + R2, we get
1 cos x
0
D= 0
1
sin + cos x = 1
0
0
1
34. Using C1 Æ C1 + C2 + C3, we get
1 a b
D = (x + a + b) 1 x b
1 b x
Using R3 Æ R3 – R2, R2 Æ R2 – R1, we get
1
a
D = (x + a + b) 0 x - a
0 b-a
b
0
x-b
= (x + a + b) (x – a) ( x – b)
35. Use R2 Æ R2 – R1, R3 Æ R3 – R1to obtain
D = sinq cosq =
1
sin(2q)
2
36. Use C1 Æ C1 + C2 + C3 to obtain
1
b
c
D = (x + a + b + c) 1 x + b
c
1
b
x+c
Now use R2 Æ R2 – R1, R3 Æ R3 – R1 to obtain
D = x2(x + a + b + c)
Thus,
D = 0 fi x = 0 or x = – (a + b + c)
37. Use C1 Æ C1 – C2.
38. Show D = – a3 + 8b3
fi a/b is a cube root of 8.
39. Using C1 Æ C1 + C2 + C3, we can write D =
2(1 + i)D1 where
1 1+ i
i
1 1+ i
i
1
i
1
D1 =
= 0 -1 1- i = i
1 1 1+ i
0 -i
1
[using R2 Æ R2 – R1 and R3 Æ R3 – R1]
Thus, D = 2(i – 1)
1 1 1
+
c a b
1 1 1
40. Write D = abc 1
+
a b c
1 1 1
1
+
b c a
and use C2 Æ C2 + C1, to show that D = 0
1
41. Applying C1 Æ C1 – C2 – C3 and reduce D to
determinant in Example 36.
42. Applying R1 Æ R1 – xR2 we get
a (1 - x 2 ) c (1 - x 2 )
D = ax + b
cx + d
u
v
Take 1 – x2 common
– xR1 to obtain
a c
2
D = (1 – x ) b d
u v
p (1 - x 2 )
px + q
w
from R1 and apply R2 Æ R2
p
q
w
43. Applying R1 Æ R1 + R3 to obtain
0
D = - sin q
-1
0
1
- sin q
2
sin q = 2(sin2q + 1)
1
As 0 £ sin2q £ 1, we get D Œ [2, 4].
44. Write D = D1 – D2 where
b2 - ab b
D1 = ab - a 2
bc - ac
a b2 - ab and
c
ab - a 2
b2 – ab c
bc – ac
bc - ac
2
b b2 – ab
bc – ac
a ab – a 2
D2 = ab – a
In D1 use C1 Æ C1 – (b – a)C2 to show that D1
= 0.
4.48
Complete Mathematics—JEE Main
In D2 use C3 Æ C3 – (b – a)C2 to show that D2
= 0.
Thus, D = 0.
45. We can write
6i 1
1
D = – 3iw 4 - 1 -1 = 0
20 i
i
\ x = 0, y = 0
46. Using C1 Æ C1 + C2 + C3, we get
49. Show that the determinant equals
x3 + (a2 + b2 + c2) x = 0.
fi x = 0 as x ΠR
\ Number of real roots of the equation is one.
50. Note that
C11 C12 C13
D1 = C21 C22 C23
C31 C32 C33
where Cij cofactor of (i, j)th element of
È1
Í
Íx
Í 2
Îx
1 cos x cos x
D = (sin x + 2cos x) 1 sin x cos x
1 cos x sin x
Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get
1
cos x
cos x
D = (sin x + 2cos x) 0 sin - cos x
0
0
0
sin x - cos x
= (sin x + 2cos x) (sin x – cos x)2
D = 0 fi tan x = – 2, tan x = 1.
p
p
£x£
, – 1 £ tan x £ 1.
As –
4
4
Thus, tan x = 1 fi x = p/4.
47. Applying C1 Æ C1 + iC2, we get
0
i
-w
0
1
w2
w - iw 2
- w2
1
x + iy =
e- p i / 4
e- p i / 3
1
e - 2p i / 3
e2p i / 3
e 2p i / 3
x – iy = ep i / 4
ep i / 3
ep i / 4
ep i / 3
1
e2p i / 3
e - 2p i / 3
e 2p i / 3
1
= e -p i / 4
e -p i / 3
\
D1 = D = 49
51. Applying C3 Æ C3 – C1, we get
sin a
D = sin b
sin g
D=
e
1
e
- pi 3
- 2p i 3
e
cos b
cos b
cos b
=0
log a
log b
log 7
2
log c
log(7 ) log(73 )
1 + cos2 q
cos2 q
D = (2 + 4sin4q)
sin 2 q
cos2 q
sin 2 q
1 + cos2 q
-1
D = 2(1 + 2sin4q)
0
sin 2 q
1
-1
D = 0 fi sin4q = – 1/2
0
e
1
Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get
0
- 2p i 3
1
1 + sin 2 q 1
= 2(1 + 2sin4q)
pi 4
= e
cos a
cos a
cos a
log(3) log(32 ) log(33 )
= (log 7) (log 3) D1
log a log b log c
where D1 =
=0
1
2
3
1
2
3
\
D =0
53. C3 Æ C3 + C1 + C2, gives
2iy = (x + iy) – (x – iy)
1
1
2
Now,
- pi 4
x2
52. Using R2 Æ R2 – R1 and R3 Æ R3 – R1, and
Ê xˆ
log x – log y = log Á ˜ , we get
Ë y¯
= (w – iw2) (iw2 + w)
= w2 – i2w4 = w2 + w = – 1
fi x = – 1, y = 0.
48. Taking conjugate, we get
1
x2 ˘
˙
1˙
˙
x˚
x
-e
= (e – 2pi/3 – e2pi/3) [1 – 1] = 0
fi y = 0.
Now, 0 £ q £
2p i 3
\
4q =
p
fi 0 £ 4q £ 2p
2
7p 11p
,
6
6
fi
q=
7p 11p
,
24 24
1
0
0
Determinants 4.49
54.
2
( x -1)( x +1)
( x +1)
(2 x +1)( x +1)
P(x) = (2 x -1)( x +1) (2 x -1)( x + 3) (2 x + 3)(2 x -1)
(2 x +1)(3 x - 2) (2 x -1)(3 x - 2) (4 x +1)(3 x - 2)
x - 1 x + 1 2x + 1
= ( x + 1)(2 x - 1)(3 x - 2) x + 1 x + 3 2 x + 3
2x + 1 2x - 1 4x + 1
Using C2 Æ C2 – C1, C3 Æ C3 – 2C1, we get
x -1 2 3
P(x) = ( x + 1)(2 x - 1)(3 x - 2) x + 1 2 1
2 x + 1 -2 -1
Using R2 Æ R2 – R1, R3 Æ R3 + R1, we get
x -1 2 3
P(x) = ( x + 1)(2 x - 1)(3 x - 2) 2
0 -2
3x 0 2
P(x) = – 4(x + 1) (2x – 1) (3x – 2) (3x + 2)
\ P(x) = 0 has four distinct roots.
2p
2p
+ i sin
.
9
9
55. Put ar = wr where w = cos
Note that w9 = 1 fi
1
D= w
3
w6
1
= w8. Now,
w
w8
w7
2
w
w5
w4
w
=0
as C2 and C3 are proportional.
56. As the system has a non-zero solution
p b c
D= a q c = 0
a b r
Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get
p
b
c
0
=0
D= a- p q-b
a- p
0
r-c
Expanding along R3, we get
– (a – p) (q – b)c + (r – c) [p(q – b) – b(a – p)]
= 0.
Dividing (p – a) (q – b) (r – c), we get
c
p
b
+
+
= 0.
r-c
p-a
q-b
fi
c-r
r
p
b-q
q
+
+
+
+
=0
r-c r–c p-a q-b q-b
p
q
r
+
+
= 2.
p-a q-b r-c
Now use
p+a
2 p - ( p - a)
2p
=
=
- 1 etc.
p-a
p-a
p-a
57. We can write D as
1 n + 1 (n + 1) (n + 2)
D = n! (n + 1)! (n + 2)! 1 n + 2 (n + 3) (n + 2)
1 n + 3 (n + 4) (n + 3)
fi
Applying R3 Æ R3 – R3, R2 Æ R2 – R1, we get
1 n + 1 (n + 1) (n + 2)
D = n! (n + 1)! (n + 2)! 0
1
2 ( n + 2)
0
1
2 (n + 3)
fi
D
=2
n!(n + 1) ! (n + 2)!
58. Use R1 Æ R1 + R3 – 2R2
59. Using C1 Æ C1 + C2 + C3, we get
1 1- y 1- y
D = (3 – y) 1 1 + y 1 - y
1 1- y 1+ y
Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get
1 1- y 1- y
D = (3 – y) 0 2 y
0 = (3 - y ) ( 4 y 2 )
0 -2 y
2y
\
D = 0 fi y = 0 or y = 3.
60. Write D = lD1 + l¢D2, where
a am + a¢ m¢ an + a ¢ n¢
D1 = b bm + b¢ m¢ bn + b¢ n¢ and
c cm + c¢ m ¢ cn + c¢ n¢
a¢ am + a ¢ m¢ an + a¢ n¢
D2 = b¢ bm + b¢ m¢ bn + b¢ n¢
c¢ cm + c¢ m¢ cn + c¢ n¢
In D1 apply C2 Æ C2 – mC1, C3 Æ C3 – nC1 to
obtain
a a ¢ m ¢ a ¢ n¢
D1 = b b ¢ m ¢ b ¢ n ¢ = 0
c c ¢ m ¢ c ¢ n¢
Similarly D2 = 0. Thus, D = 0
61. Imitate Example 35.
62. We have
4.50
D=
Complete Mathematics—JEE Main
1
m
1
m+3
1
m+6
1
1
1
(m + 3) (m + 2)
(m + 6) (m + 5)
m (m - 1)
2
2
2
Applying C3 Æ C3 – C2, C2 Æ C2 – C1, we get
1
m
D=
0
3
0
3
1
m (m - 1) 3 (m + 1) 3(m + 4)
2
2
Using C1 Æ C1 + C2 + C3, we get
1
0
cos C cos B
1
64. Write D =
- 1 cos A
a a cos C
a cos B cos A - 1
Applying C1 Æ C1 + bC2 + cC3, we get
a b cˆ
Ê
D = pqr Á 1 + + + ˜
Ë
p q r¯
\
lim
x Æ0
x2
1 1 1
1 0 0
= 2 1 2 = 2 1 2 =–1
0 1 1
0 1 1
a
b
aa +b
D=
b
c
ba + c = 0
aa + b ba + c
0
Applying R3 Æ R3 – aR1 – R2 to obtain
66. Applying C2 Æ C2 – w2C1 we find the second
column of D becomes 0.
67. The system will have a non-trivial solution if
p+a
b
c
q+b
c
=0
b
r+c
a
a
0 0 – (a a 2 + 2ba + c)
-1
D = pqr
a
p
a
p
a
p
b
q
1+
b
q
b
q
c
r
c
r
1+
c
r
c
c
Using C3 Æ C3 – C2, C2 Æ C2 – C1, we get
a
0
- (1 + a)
D = b - (1 + b) (b + 1)
-1
c +1
0
= – a( b + 1) (c + 1) – (1 + a)[b(c + 1)
– (1 + b)]
1
a
b
=0
\
D=0fi +
1+ a 1+ b 1+ c
1
1
1
= 2.
+
+
1+ a 1+ b 1+ c
70. The system will have a non-trivial solution if
D = 0 where
l+5 l-4 1
D= l -2 l +3 1
l
l
1
Applying C2 Æ C2 – lC3, C1 Æ C1 – lC3, we get
fi
Write
1+
aa +b
ba + c
a b
D= b c
[using R1 Æ R1 – R3]
D=
a b c
+ + = – 1.
p q r
= – (aa2 + 2ba + c) (ac – b2)
Note that D = 0 if a, b, c are in G.P.
69. The system will have a non-trivial solution if
a
a -1
D= b -1
b = 0.
cos x 1 1
sin x
f ( x)
= 2
1 2
2
x
x
tan x 1 1
f ( x)
D = 0, pqr π 0 fi
(1)
68. The system will have a non-zero solution if
a cos C cos B
1
D=
0 - 1 cos A = sin2A
a
0 cos A - 1
65. For x π 0,
c
r
c
r
a b cˆ
b
Ê
D = pqr Á 1 + + + ˜ 1 1 +
Ë
p q r¯
q
b
c
1
1+
q
r
Apply R2 Æ R2 – R1, R3 Æ R3 – R1, we get
3
= 3 (m + 4 – m – 1) = 3
\ a+b+g=3
63. Use C2 Æ C2 – 2C1, C3 Æ C3 – 3C1 to obtain C2
and C3 are proportional.
b
q
Determinants 4.51
5 -4 1
D= -2 3 1 = 7 π 0
0
0 1
Thus, the system cannot have a non-trivial solution.
71. Adding above the system of equations, we get
(a + b + c) (x + y + z) = 0 fi x + y + z = 0
\ (b + c) (– x) – ax = b – c
b-c
etc.
fi x=–
a+b+c
72. The above system of equations will have a nontrivial solution if
a b c
D= b c a = 0
c
But D = –
1
(a + b + c)
2
[(b – c)2 + (c – a)2 + (a – b)2]
6 6
1 -6
0
4
p2
p3
p
77. Statement-2 is true, see theory.
The roots of x9 = p are p1/9 wr
2p
2p
where w = cos
+ i sin
.
9
9
value of determinant D depends on a1, a2, . . . a9.
If we put ak = p1/9 w k, then
1 w8
a1
D1 = a2
a3
b1 + q c1 c1 + ra1
b2 + qc2 c2 + ra2
b3 + qc3 c3 + ra3
b1
D2 = b2
b3
b1 + q c1 c1 + ra1
b2 + qc2 c2 + ra2
b3 + qc3 c3 + ra3
w7
D = (p1/9)3 1
w
w2 = 0
1
w
w2
However, if a1 = p1/9, a2 = p1/9 w8,
a3 = p1/9 w7, a4 = p1/9 w, a5 = p1/9 w5
a6 = p1/9 w4, a7 = p1/9 w2, a8 = p1/9 w3,
a9 = p1/9 w6, then
1 w8 w7
D = (p1/9)3 w
w2
\
f ¢¢¢(x) is a constant.
74. Write
D = D1 + pD2 where
w5 w4
w3
w6
= p1/3 (2w2 – w7 – w6) π 0.
78. Statement-2 is true.
and
Now show that D1 = D and D2 = pqr D.
75. The system will have a non-trivial solution if
l +3 l +2 1
D=
3
l + 3 1 = 0.
2
3
1
Using R1 Æ R1 – R2 and R2 Æ R2 – R3, we get
l
D= 1
2
[using statement-2]
a b
\
D = 0, a + b + c π 0
fi
a = b = c.
Thus, a : b : c = 1 : 1 : 1
73. f ¢¢¢(x) =
If n is odd, then |A| = – |A| fi 2|A| = 0 fi |A| = 0.
We have
0
sin(p / 12) cos(p / 12)
Êpˆ
D Á ˜ = - sin(p / 12)
0
sec(p / 12) = 0
Ë 4¯
- cos(p / 12) - sec(p / 12)
0
-1 0
2
l 0 =l +1
3 1
Note that there is no real value of l for which
D = 0.
76. Let A be a skew symmetric matrix of order n, then
|A¢| = (– 1)n |A| fi |A| = (– 1)n |A|
Using R2 Æ R2 –
get
6
w=
2 R1 and R3 Æ R3 –
2i
3 R1, we
3 + 6i
0
3 ( 6 - 2 3 )i
0
2
(2 - 3 2 )i
6 ÎÈ2 3 – 3 6 – 2 3 + 2 6 ˘˚ i = – 6i
79. We first show statement-2 is true.
Ê nˆ
Ê nˆ
Ê nˆ
P(x) = 1 + Á ˜ x + Á ˜ x 2 + . . . + Á ˜ x n
Ë1 ¯
Ë 2¯
Ë n¯
=
fi
Ê nˆ
Ê nˆ
Ê nˆ
P¢(x) = Á ˜ + 2 Á ˜ x + . . . + n Á ˜ x n - 1
Ë1 ¯
Ë 2¯
Ë n¯
Ê nˆ
P¢(0) = Á ˜ = coefficient of x in the expanË1 ¯
sion of P(x)
\
Statement-2 is true.
Note that D(x) consists of 6 terms of the form
(1 + x)n.
Thus, coefficients of x in D(x) = D¢(0)
a1b1 a1b2 a1b3
1
1
1
But D¢(0) = 1
1
1 + a2 b1 a2 b2 a2 b3
1
1
1
1
1
1
fi
4.52
Complete Mathematics—JEE Main
1
1
1
+ 1
1
1 =0
a3 b1 a3 b2 a3 b3
80. Statement-2 is not always true. For instance, the
system of equations
x + 2y + 3z = 1
2x + 3y + 4z = 2
3x + 4y + 5z = 4
has no solution but |A| = 0
For statement-1, let
a b c
D= b c a
c a b
Using C1 Æ C1 + C2 + C3, we obtain
1 b c
D = (a + b + c) 1 c a = 0
1 a b
Also, note that x = y = z satisfies each of the three
equations.
Thus, the system of equations has infinite number
of solutions.
81. Let
a b c
D= b c a
c
a b
1
= - (a + b + c)[(b - c)2 + (c - a)2 + (a - b)2 ]
2
If a + b + c π 0 and a2 + b2 + c2 = bc + ca + ab,
then (b – c)2 + (c – a)2 + (a – b)2 = 0
fi a=b=cπ0
[ a + b + c π 0]
Thus, the system of equation reduces to x + y + z = 0
which is satisfied by infinite number of solutions.
\ Statement-1 is true.
If a + b + c = 0 and a2 + b2 + c2 π bc + ca +
ab, then at least one following is true:
b2 π ca,
c2 π ab.
a2 π bc,
Suppose b2– ac π 0
Write first two equations as:
ax + by = (a + b)z
bx + cy = (b + c)z
Eliminating y, we get
(ac – b2)x = (ac – b2)z
fi
x= z
Thus, from (1) equation, we get
by = bz fi y = z
\
x= y=z
Therefore, statement-2 is false.
82. If (a1, b1), (a2, b2) (a3, b3) are non-collinear, then
D1 π 0. Therefore statement-2 is true.
We can write D2 as
a1 b1 0 b1 a1 0
D2 = a2 b2 0 b2 a2 0 = 0
a3 b3 0 a3 b3 0
for all values of a1, a2, a3, b1, b2, b3 ΠR.
a + ib c + id
83. As
-c + id a - ib
= (a + ib) (a – ib) + (c – id) (c + id)
= a2 + b2 + c2 + d2
we get statement-2 is true.
Now,
(a12 + b12 + c12 + d12 ) (a22 + b22 + c22 + d22 )
=
a1 + ib1
-c1 + id1
c1 + id1 a2 + ib2
a1 - ib1 - c2 + id2
c2 + id2
a2 - ib2
a + ib c + id
= a 2 + b2 + c2 + d 2
-c + id a - ib
where a = a1a2 – b1b2 + c1c2 – d1d2,
b = a 2b 1 + a 1b 2 + c 1d 2 + c 2d 1,
c = – a 2c 1 + b 2d 1 – b 2c 1 + a 2d 1,
d = – b 2c 1 + a 2d 1 – b 1c 2 – b 1d 2
=
Level 2
84. We have
l1 m1
2
D = l2 m2
l3 m3
a1
= b12
b13
b21
a2
b23
n1
n2
n3
l1
l2
l3
m1
m2
m3
n1
n2
n3
b31
b32
a3
where ak = lk2 + mk2 + nk2 = 1 for k = 1, 2, 3
and bij = li lj + mimj + ninj = 0 " i π j
1 0 0
Thus, D = 0 1 0 = 1 fi |D| = 1.
0 0 1
85. Using R1 Æ aR1, R2 Æ bR2, R3 Æ cR3,
we get
2
ab2 c 2
abc ab + ac
1
2 2
a bc abc bc + ab
D=
abc 2 2
a b c abc ac + bc
bc 1 ab + ac
a2 b2 c 2
=
ca 1 bc + ab
abc
ab 1 ac + bc
Determinants 4.53
using C3 Æ C3 + C1, we get C2 and C3 are
propotional. Thus, D = 0
86. Using C1 Æ C1 – (x2/3)C3, C2 Æ C2 – (2x/3) C3,
we get
- 5x + 3 - 5 3
D(x) =
x+9
1 9
- 6 x + 9 - 6 21
Using C1 Æ C1 – xC2, we get
3 -5 3
D(x) = 9 1 9 = a constant.
9 -6 21
Thus, a = 0.
Alternate Solution
Replacing x by 1/x, we get
1
Ê 1ˆ
Ê 1ˆ
-5 Á ˜ + 3 2 Á ˜ - 5
2
Ë x¯
Ë x¯
x
Ê 1ˆ
3Á 2 ˜
Ëx ¯
Ê 1ˆ
= aÁ 3˜
Ëx ¯
1 - 5x + 3x
fi 3 + x + 9x
2
7 - 6x + 9x
Putting
1
a= 3
7
x=
2
6
14
2
2 - 5x
6+x
b2 (1 - cos q )
a2 (1 - cos q )
b2 + (c 2 + a2 )cos q
a2 (1 - cos q )
b2 (1 - cos q )
D=
c 2 (1 - cos q )
c 2 + (a2 + b2 )cos q
Using C1 Æ C1 + C2 + C3 and a2 + b2 + c2 = 1,
we get
b 2 (1 - cos q )
1
2
2
D = 1 b + (1 - b ) cos q
3
b 2 (1 - cos q )
1
1
Ê 1ˆ
+9
6Á ˜ +1 9
Ë x¯
x
Ê 1ˆ
Ê 1ˆ
-6 Á ˜ + 9 14 Á ˜ - 6 21
Ë x¯
Ë x¯
Ê 1ˆ
Ê 1ˆ
+ bÁ 2 ˜ + cÁ ˜ + d
Ëx ¯
Ë x¯
2
a2 + (b2 + c 2 )cos q
c 2 (1 - cos q )
c 2 (1 - cos q )
c 2 (1 - cos q )
c 2 + (1 - c 2 ) cos q
Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get
+
Ê 1ˆ
7Á 2 ˜
Ëx ¯
88. First multiply C1 by a, C2 by b, C3 by c, followed
by multiplying R1 by 1/a, R2 by 1/b and R3 by 1/c,
we get
1 b2 (1 - cos q ) c 2 (1 - cos q )
cos q
D= 0
0
= cos2 q
cos q
0
0
89. We have
- sin(a + q ) sin(a + q ) 1
3
2
9 = a + bx + cx + dx
3
14 - 6 x 21
0, we get
3
9 =0
21
[ C1 and C2 are proportional]
87. The given system of equations will have a nontrivial solution if
a+t
b
c
D= a
b+t
c =0
a
b
c+t
Using C1 Æ C1 + C2 + C3, we get
1 b
c
c =0
D = (a + b + c + t) 1 b + t
1 b
c+t
Using C2 Æ C2 – bC1, C3 Æ C3 – cC1, we get
D = (a + b + c + t)t2 = 0 fi t = 0, – (a + b + c)
Thus, there are just two distinct values of t.
dA
= - sin(b + q ) sin(b + q ) 1
dq
- sin(g + q ) sin(g + q ) 1
cos(a + q ) cos(a + q ) 1
+ cos(b + q ) cos(b + q ) 1
cos(g + q ) cos(g + q ) 1
=0+0=0
fi A is independent q. Thus, A(a, b, g, q ) =
A (a, b, g , 0 )
p 2p ˆ
p ˆ
Ê p
Ê p
fi A Á - , 0, , ˜ = A Á - , 0, , 0˜
Ë 2
¯
Ë
2 13
2
2 ¯
0 -1 1
= 1 0 1 =2
0 1 1
90. We have D = (a – b) (b – c) (c – a) = – 2.
As a > b > c, a – b, b – c are positive integers and
c – a is a negative integers.
Only possibilities are
a – b = 2, b – c = 1, c – a = –1
(1)
4.54
Complete Mathematics—JEE Main
or a – b = 1, b – c = 2, c – a = –1
(2)
or a – b = 1, b – c = 1, c – a = –2
(3)
(1) and (2) lead us to 0 = 2.
\ a – b = 1, b – c = 1, c – a = –2.
Now, 3a + 7b – 10c = 3(a – c) + 7(b – c) = 13
91. Write
cos A sin A 0
D = cos B sin B 0
cos C sin C 0
cos P - sin P 0
cos Q - sin Q 0
cos R - sin R 0
=0
93. For each D ΠB, there exist D1 ΠC where D1 is
obtained by interchanging 1st and 2nd row of D,
similarly, for each D ΠC there exists D1 ΠB.
\ B and C has same number of elements.
94. |a|2 = |x|2 + |y|2 + |z|2 + x y + x y + x z + x z
+ yz + yz
\ |a|2 + |b|2 + |c|2 = 3(|x|2 + | y|2 + |z|2) +
( x y + x y + x z + x z + y z + y z) (1 + w + w2)
fi
| a |2 + | b |2 + | c |2
95. Write the determinant as = abc D1 – D2
Alternate Solution
1 a a2
Write D = cos P D1 – sin P D2
cos A cos( A + Q) cos( A + R)
where D1 = cos B cos( B + Q) cos( B + R)
cos C cos(C + Q) cos(C + R)
sin A cos( A + Q) cos( A + R)
and D2 = sin B cos( B + Q) cos( B + R)
sin C cos(C + Q) cos(C + R)
In D1, use C2 Æ C2 – cosQ C1
C3 Æ C3 – cosR C1
=0
[
C2 and C3 are propotional]
Similarly, D2 = 0
D=0
92. Expanding along C1, we get
f(x) = 2cosx
where D1 = 1 b b2
1 c c
= (a – b) (b – c) (c – a)
2
a a2 1
and D2 = b b2 1 = D1
c c2
1
\ (a – b) (b – c) (c – a) (abc – 1) = 0
Since, a, b, c are distinct, we get
abc – 1 = 0 or abc = 1
cos A sin A sin Q sin A sin R
D1 = cos B sin B sin Q sin B sin R
cos C sin C sin Q sin C sin R
Thus,
=3
| x |2 + | y |2 + | z |2
2 cos x
1
1
0
–
1
2 cos x
1 2 cos x
1 4 4
96. | P | = |adj P| = 2 1 7 = 4
1 1 3
2
fi |P| = ± 2
97. Using R2 Æ R2 + 5R1, R3 Æ R3 – 4R1,
we get
1
-3
4
0 x - 13
22
0
13
x - 22
=0
= 8cos3 x – 4cos x = 4cos x cos2x
fi (x – 13) (x – 22) – (13)(22) = 0
Êpˆ
Êpˆ
fi f Á ˜ = –1, f Á ˜ = 0.
Ë 2¯
Ë 3¯
fi x(x – 35) = 0 fi x = 0, 35
f¢(x) = – 4[sinx cos2x + 2cosx sin2x]
= – 4[sin3 x + cosx sin2x]
Êpˆ
È
Êpˆ
Ê 2p ˆ ˘
fi f ¢ Á ˜ = - 4 Ísin p + cos Á ˜ sin Á ˜ ˙
¯
Ë 3 ¯˚
Ë 3¯
Ë
3
Î
=–
3
Previous Years’ AIEEE/JEE Main Questions
1. As ax2 + 2bx + c = 0 has equal roots,
b2 – ac = 0
Using R3 Æ R3 – xR1 – R2, we get
a b ax + b
D = b c bx + c
0 0
a
Determinants 4.55
where a = 0 – x(ax + b) – (bx + c)
= – (ax2 + 2bx + c)
\
fi
D = a (ac – b2) = 0
2. As l, m, n are pth, qth, rth terms of a G.P., log l,
log m, log n are pth, qth, rth terms of an A.P. Let
a and its common difference be d.
Now,
a + ( p - 1)d
D = a + (q - 1)d
a + (r - 1)d
p 1
q 1
r 1
Using C1 Æ C1 – dC2 – (a – d)C3, we get
0 p 1
D= 0 q 1=0
0 r 1
3. If n is a multiple of 3, we get each element of D
becomes 1.
\
D=0
If
n = 3k + 1, then
a, c, b are in H.P.
5. Let R be the common ratio of the G.P. Applying
C3 Æ C3 – C2 and C2 Æ C2 – C1, we get that the
given determinant
log an
= log an +1
log an + 2
log R log R
log R log R = 0
log R log R
6. Applying C1 Æ C1 + C2 + C3, we get
1 + 2 x + (a 2 + b 2 + c 2 ) x (1 + b 2 ) x (1 + c 2 ) x
f(x) = 1 + 2 x + (a 2 + b 2 + c 2 ) x
1 + b2 x
1 + 2 x + (a 2 + b 2 + c 2 ) x (1 + b 2 ) x
(1 + c 2 )xx
1 + c2 x
1 (1 + b 2 ) x (1 + c 2 ) x
=1
1 + b2 x
(1 + c 2 ) x
1 (1 + b 2 ) x
1 + c2 x
Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get
1
w
D= w
w2
w2
1
w2
1 =0
[use C1 Æ C1 + C2 + C3]
1 (1 + b 2 ) x (1 + c 2 ) x
f ( x) = 0
1- x
0
0
0
1- x
= (1 – x)2
w
which is a polynomial of degree 2.
If n = 3k + 2, then
1
w2
w
D = w2
w
1 =0
w
1
7. If A
[use C1 Æ C1 + C2 + C3]
w2
4. As the system has a non-zero solution
1 2a a
1 3b b = 0
1 4c c
1/ a 2 1
1/ b 3 1 = 0
1/ c 4 1
Using R1 Æ R1 – 2R2 + R3, we get
fi
1 2 1
- + =0
a b c
1/ a - 2 / b + 1/ c 0 0
1/ b
3 1 =0
1/ c
4 1
a 1 1
1 1 1
| A | = 1 a 1 = (a + 2) 1 a 1
1 1 a
1 1 a
[using C1 Æ C1 + C2 + C3]
1
1
1
= (a + 2) 0 a - 1
0
0
0
a -1
= (a + 2)(a – 1)2
If |A| π 0, the system has a unique solution.
If |A| = 0, then a = – 2 or a = 1. For a = 1, the
system of equations becomes x + y + z = 0 which
For a = – 2, the system of equation becomes
– 2x + y + z = – 3
x – 2y + z = – 3
4.56
Complete Mathematics—JEE Main
Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get
x + y – 2z = – 3
which on adding becomes 0 = – 9
Thus, the system has no solution if a = – 2.
1 1
1
8. D = 1 1 + x
1
1 1 1+ y
1
b
c
D1 = 0 c - b a - c
0 a-b b-c
= – (b – c)2 – (a – b)(a – c)
= – (a2 + b2 + c2 – bc – ca – ab)
1
D1 = - [(b - c)2 + (c - a)2 + (a - b)2 ] < 0
2
As a + b + c > 0, we get
Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get
fi
1 1 1
D = 0 x 0 = xy
0 0 y
D = (a + b + c) D1 < 0
11. Using C1 Æ C1 – C3 in D, we get
which is divisible by both x and y.
9. Let D 2 =
a +1
a -1
b +1
b -1
0
D= 0
2 cos a
c -1
c +1
( -1)n + 2 a ( -1)n +1 b ( -1)n c
sin a
cos a
- sin a
cos a
sin a
- cos a
= 2 cos a (sin2 a – cos2 a)
a + 1 a - 1 ( -1)n + 2 a
= –2 cos a cos 2a
D 2 = b + 1 b - 1 ( -1)n +1 b
c -1 c +1
D = 0 for a = p/4 Π(0, p/2).
( -1)n c
This is the only value of a lying in (0, p/2) for
which D = 0.
(-1) n+2 a a + 1 a - 1
= (-1) 2 (-1) n+1 b
(-1) n c
The system of linear equation will have a non-trivial
solution if and only if
b +1 b -1
c -1 c +1
Thus,
n+ 2
a a + 1 a - 1 ( -1) a a + 1 a - 1
D = -b b + 1 b - 1 + ( -1)n+1 b b + 1 b - 1
c c -1 c +1
( -1)n c c - 1 c + 1
a + ( -1)n + 2 a
a +1 a -1
n +1
b b +1 b -1
c + ( -1)n c
c -1 c +1
= -b + ( -1)
1 sin a
D1 = 1 cos a
1 - sin a
cos a
sin a = 0
- cos a
Using R2 Æ R2 + R1, we get
2
0
D1 = 1 cos a
1 - sin a
fi
fi
0
sin a = 0
- cos a
2(– cos2 a + sin2 a) = 0
– 2 cos 2a = 0
This is true for only one value of a Π(0, p/2)
n is any odd
integer.
\
D = 0 if n is any odd integer.
10. Using C1 Æ C1 + C2 + C3, we get
D = (a + b + c) D1
where
1 b c
D1 = 1 c a
1 a b
viz, a = p/4.
Thus, statement-1 is also true. However statement-2
is not a correct reason for statement-1.
1+1+1
12. D = 1 + a + b
1+ a2 + b2
1+ a + b
1+ a2 + b2
1+ a2 + b2
1 + a3 + b3
1 + a3 + b3 1 + a 4 + b4
Determinants 4.57
1
=1
1
a
1 a
2
1 1
b 1
1
a
2
b 1 a
1
where D1 = 1
1
a
we get
1
b = D12
2
b
b1 x + c1
| A( x ) | = 2 b2 x + c2
b3 + c3
2
1
b
2
Applying C1 Æ C1 – xC2, we get
0
a -1
c1
| A( x ) | = 2 c2
c3
0
b -a
\
1 a2 -1 b2 - a2
= (a - 1)(b - a )
1
1
a +1 b +a
\
b1
b2
b3
a1
a2
a3
|B(x)| is independent of x.
15. Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get
a2
D = 2 al + l 2
-4al
= (a – 1) (b – 1) (b – a)
Thus,
a1
a2
a3
2
1 a
b
Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we get
1
D1 = 1
b1
b2
b3
D = (a – 1)2 (b – 1)2(a – b)2
b2
c2
2 bl + l 2
-4bl
2cl + l 2
-4cl
Take – 4l common from R3, and applying
R2 Æ R2 – 2lR3, we get
K=1
13. As the given system of equations has a non-trivial
solutions.
a2
D = -4l 3 1
a
a - 1 -1
-1
D = -1 b - 1 -1 = 0
-1
-1 c - 1
a2
= l (4l 2 ) a
1
Write
a -1
-1
1 -1
-1
D = 0 b - 1 -1 - 1 b - 1 -1
0 -1 c - 1 1 -1 c - 1
1 0 0
= a[(b - 1)(c - 1) - 1] - 1 b 0
1 0 c
b2
1
b
\
c2
1
c
b2
b
1
c2
c
1
k = 4l2
n -1
16. As  r =
r =1
n -1
1
n(n - 1), Â (2r - 1) = (n - 1)2
2
r =1
n -1
and  (3r - 2) =
r =1
[use C2 Æ C2 + C1, C3 æ C3 + C1]
=
= a (bc – b – c) – bc
3
n(n - 1) - 2(n - 1)
2
1
(n - 1)(3n - 4)
2
Thus,
As D = 0, we get
ab + bc + ca = abc
n(n - 1)/ 2 (n - 1)2
n -1
 Dr = n / 2
n -1
14. We have |B(x)| = |(A(x))TA(x)|
T
r =1
n(n - 1)/ 2 (n - 1)2
(n - 1)(3n - 4)/ 2
a
(n - 1)(3n + 4)/ 2
= |A(x) | |A(x)|
= |A(x)| |A(x)| = |A(x)|2
Let pi(x) = ai x2 + bi x + ci (i = 1, 2, 3)
Applying C1 Æ C1 -
1 2
x C3 , C2 Æ C2 - xC3 .
2
Using R3 Æ R3 – R1, we get
n(n - 1)/ 2 (n - 1)2
n -1
 Dr = n / 2
r =1
0
0
n -1
(n - 1)(3n - 4)/ 2
a
4(n - 1)
4.58
Complete Mathematics—JEE Main
Expanding along R3, we get
n -1
 D r = 4(n - 1)
r =1
n(n - 1)/ 2 (n - 1)2
n/2
n -1
2 2 -1
4 3 0 = a - 12
6 1 1
fi 12 = a – 12
n -1 n -1
Ê nˆ
= 4(n - 1) Á ˜ (n - 1)
=0
Ë 2¯
1
1
17. The system will have a non-trivial solution if
2-l
D= 2
-1
-2
-3 - l
2
1
2 =0
-l
Using R1 Æ R1 + R3 , R2 Æ R2 + 2R3,
1- l
0
1- l
D= 0
1 - l 2 - 2l = 0
-1
2
-l
1 0 1
(1 - l ) 0 1 2 = 0
-1 2 - l
2
fi
Using C3 Æ C3 – C1, we get
1 0
0
2
(1 - l ) 0 1
2 =0
-1 2 - l + 1
fi a = 24
19. Let x be any real number and y = 1, z = 1, then
x 1 1
D= 1 1 1 =0≥0
1 1 1
and xyz = x can take any real value, thus, minimum
value of xyz does not exist.
20. The given system of equations has a non-trivial
solution if
1 l -1
l -1 -1 = 0
1 1 -l
fi l – l – l – 1 + 1 + l3 = 0
fi l (l – 1) (l + 1) = 0
fi l = 0, –1, 1
When l = 0, x = –y = z = k
0
When l = –1, x = 0, y = –z = k
When l = 1, y = 0, x = –z = k
0
0
2
fi (1 – l) [ – l + 1 – 4] = 0 fi l = 1, – 3
Thus, the set contains two elements
18. Using R2 Æ R2 – R1 – R3, we get
x2 + x
-4
D=
x +1
0
x-2
0
x2 + 2 x + 3 2 x - 1 2 x - 1
Thus, there are three values of l, for which the
system of equations has a non-trivial solutions.
21. Using C1 Æ C1 + C2 + C3, we get
1 sin x sin x
D = (2 sin x + cos x) 1 cos x sin x
1 sin x cos x
Using C2 Æ C2 – (sin x) C1
Expanding along R2, we get
and C3 Æ C3 – (sin x) C1,
x +1 x - 2
2x - 1 2x - 1
Using C2 Æ C2 – C1, we get
we get
D = -( -4)
D=4
x + 1 -3
2x - 1 0
= 4(3) (2x – 1)
= 24x – 12
1
0
0
D = (2 sin x + cos x) 1 cos x - sin x
0
1
0
cos x - sin x
= (2 sin x + cos x) (cos x – sin x)2
1
\ D = 0 fi tan x = – , 1
2
As –p/4 £ x £ p/4, –1 £ tan x £ 1
\ a = 24
and tan x is one-to-one in the interval [– p/4, p/4].
TIP : Put x = 1 to obtain
Thus, there are two values of x.
Determinants 4.59
Thus, the system has non-trivial solution for no
value of l.
Previous Years’ B-Architecture Entrance
Examination Questions
a
a
2
3. Â D k =
(b – 1)x = 0
k =1
n
In this case y + z = 0, by + z = 0
where a = Â 2(3k -1 )
fiy=0
(b – 1)y = 0
k =1
Thus, z = 0
2(3n - 1)
= 3n - 1,
3 -1
=
That is, if b π 1, then the system has only trivial
solution.
\
b=1
n
b = Â 3(4 k -1 ) =
k =1
3(4 n - 1)
= 4n - 1
4 -1
n
and c = Â 4(5k -1 ) =
(a – 1) x = 0.
k =1
If a π 1, x = 0 and y + z = 0
a
Thus, Â D k = a
k =1
a
n
tion.
If a = 1, all the three equations become identical to
x + y + z
of solutions.
b = 1 fi (b – 1)2 = 0
2 -1 a12
2 -2 a13
4. det(B) = 2a21
a22
2 -1 a23
22 a31
2a32
a33
a11
fi b2 = 2b – 1
22 a11
l +3 l +2 1
D= 3
l +3 1 =0
2
3
1
n
5. Â D r =
r =1
a12
a22
a32
2
a33
a13
a23
a33
a
a
b
b
c
g
2 n - 1 3n - 1 5n - 1
n
where
4x + 3y + z = 0
(i)
3x + 4y + z = 0
(ii)
2x + 3y + z = 0
(iii)
a = Â 2r -1 = 2 n - 1
r =1
n
b = Â 2(3r -1 ) =
r =1
From (i) and (iii) x = 0
y = 0, z = 0
2a32
= det (A) = 5
For l = 1, the system becomes
fi
a23
a11
= (2 )(2 )(2) a21
a31
fil=1
4y + z = 0
= (2 -2 )(2 -1 ) 22 a21 2a22
-3
fi D = l(l + 1) – (l – 1) = 0
and
a13
2 a31
l +1 l -1 0
D= 1
l
0 =0
2
3
1
From (i) and (ii) 3y + z = 0
2a12
2
Using R1 Æ R1 – R3, R2 Æ R2 – R3, we get
fi (l – 1) = 0
4(5n - 1)
= 5n - 1
5 -1
b c
b g =0
b c
2. As the system has a non-trivial solution,
2
c
g
3n - 1 4 n - 1 5n - 1
If b π 1, x = 0.
fi
b
b
2(3n - 1)
3 -1
= 3n – 1
n
and
c = Â 4(5r -1 ) =
r =1
= 5n – 1
4(5n - 1)
5 -1
4.60
Complete Mathematics—JEE Main
a
 Dr = a
r =1
a
n
Thus,
b c
b g =0
b c
x1
x4
6. Taking e common from R1, e common from R2
and ex7 from R3, we get
fi
1 2 1
abc ÊÁ - + ˆ˜ ( -1) = 0
Ë a b c¯
fi
2ac = ab + bc
9. Using R1 Æ R1 – R2, R3 Æ R3 – R2, we get
0 a-c b-a
1
c
a =0
0 b-c c-a
D = e x 1 + x 4 + x 7D 1
where
1 e3 d
e6 d
fi
(a – c)2 + (b – a) (b – c) = 0
e6 d = 0
fi
a2 – 2ac + c2 + b2 – ab – bc + ac = 0
fi
a2 + b2 + c2 – bc – ca – ab = 0
where d is common difference of A.P.
fi
(a – b)2 + (b – c)2 + (c – a)2 = 0
\D=0
fi
a=b=c
fi
sin A = sin B = sin C =
D1 = 1 e3d
1 e
3d
e
6d
5 common from C2 and C3, we get
7. Taking
13 + 3
where D1 =
2
x = 2.
1
15 + 26
5
2
\
y – z = 1 and – 4y + 4z = a – 2
3 + 65
3
5
fi
4y – 4z + (– 4y + 4z) = 4 + (a – 2)
fi
0=a+2
Applying, C1 Æ C1 - 3C2 - 13C3 , we get
D1 =
3 2
Thus, sin2 A + sin2 B + sin2 C = 9/4
D = 5D1
- 3
fiA=B=C
2
1
0
5
2
0
3
5
= - 3 (5 - 6 )
Thus, D = 5 3 ( 6 - 5)
8. As the given system has a non-zero solution
1 2a a
1 3b b = 0
1 4c c
1/ a 2 1
fi abc 1/ b 3 1 = 0
1/ c 4 1
Thus, the system has no solution for a π –2.
If a
number of solutions, since y – z
number of solutions.
11. Using C1 Æ C1 + C3, we get
1 tan q + sec2 q
D= 0
cos q
0
-4
3
sin q
3
= 3 cos q + 4 sin q
dD
= – 3 cos q + 4 sin q
dq
dD
sin q cos q 1
=0fi
=
=
dq
4
3
5
fi sin q = 4/5, cos q = 3/5
Max D = max {D(0), D(sin–1 (4/5)), D(1)}
Using R1 Æ R1 – 2R2 + R3, we get
1 2 1
- +
0 0
a b c
abc 1/ b
3 1 =0
1/ c
4 1
Ï 24 ¸ 24
= max Ì3, , 4˝ =
Ó 5 ˛ 5
min D = 3.
\ D Œ[3, 5]
CHAPTER FIVE
Matrices
THE ALGEBRA OF MATRICES
A matrix is a rectangular array of numbers. The numbers
may be real or complex. It may be represented as
È a11 a12 a1n ˘
Ía
a22 a2 n ˙
˙
A = Í 21
Í ˙
Í
˙
Î am1 am 2 amn ˚
or as
A = (aij)m ¥ n.
A matrix with m rows and n columns is called as m ¥ n
matrix and the size (or dimension) of this matrix is said be
m ¥ n.
Two matrices are said to be equal provided they are of
the same size and corresponding elements are equal. For
example
È-1 2 5 ˘
Èa b c ˘
Íd e f ˙ = Í 7 3 11˙
Î
˚
Î
˚
if and only if
a = – 1, b = 2, c = 5, d = 7, e = 3 and f = 11.
Definitions A matrix A = (aij ) m ¥ n is said to be a
(i) square matrix
if m = n
(ii) row matrix
if m = 1
(iii) column matrix
if n = 1
(iv) null or zero matrix
if aij = 0 " i and j
(v) diagonal matrix
if m = n and aij = 0
"iπj
(vi) Scalar matrix
if m = n and aij = 0
" i π j and
aii = l "i
(vii) Unit or identity matrix if m = n and aij = 0
" i π j and
aii = 1 " i
(viii) Upper triangular matrix if m = n and aij = 0
"i>j
(ix) Lower triangular matrix if m = n and aij = 0
"i<j
A matrix is said to be triangular if it is either lower or
upper triangular matrix.
Addition of Matrices Two matrices A and B can be added
if and only if they are of the same size. If two matrices are
of the same size, addition is carried out term by term. For
instance
È a b c ˘ È p q r ˘ Èa + p b + q c + r ˘
Íd e f ˙ + Í s t u ˙ = Í d + s e + t f + u ˙
Î
˚ Î
˚ Î
˚
Addition is not defined for matrices of different sizes.
The additive inverse of a matrix A, denoted by – A, is the
matrix whose elements are the negatives of the corresponding elements of A. For example,
Èa b ˘
È- a -b ˘
Í
˙
– Í c d ˙ = ÍÍ -c - d ˙˙
ÍÎ e f ˙˚
ÍÎ -e - f ˙˚
If A and B are two matrices of the same size, then the
differences between A and B is defined by
A – B = A + (– B)
Thus, the subtraction is carried out term-by-term. For
instance
È a b c ˘ È p q r ˘ Èa - p b - q c - r ˘
Íd e f ˙ – Í s t u ˙ = Í d - s e - t f - u ˙
Î
˚ Î
˚ Î
˚
Properties of Addition If A, B and C are three matrices of
the same size, then
A+B=B+A
[Commutative law]
(A + B ) + C = A + (B + C) [Associative law]
A+O=O+A
[Additive property of zero]
A + ( – A) = O
where O is the null or zero matrix of the same size as that
of A.
Scalar Multiplication If A is a matrix and a is a scalar,
then aA is defined as the matrix obtained by multiplying
every element of A by a. For example
È1 - 2 5˘
È 3 - 6 15˘
= Í
3 Í
˙
˙
Î7 8 1 ˚
Î21 24 3 ˚
Properties of Scalar Multiplication If A and B are two
matrices of the same size, and a, b are two scalars, then
5.2
Complete Mathematics—JEE Main
(a + b)A = aA + bA
(ab )A = a(bA)
a(A + B) = aA + aB
Matrix Multiplication Let A = (aij)m ¥ n and B = (bij)r ¥ s be
two matrices. We say that A and B are comparable for the
product AB if n = r, that is, if the number of columns of A
is equal to the number of rows of B.
Definition: Let A = (aij)m ¥ n and B = (bij)n ¥ p be two matrices.
Their product AB is the matrix C = (cij)m ¥ p such that cij = ai1 b1j
+ ai2 b2j + ai3 b3j + ... + ain bnj for l £ i £ m, 1 £ j £ p. Note that
cij, the (i, j)th element of AB, has been obtained by multiplying
ith row of A, namely (ai1 ai2 ai3 º ain)
Ê b1 j ˆ
Áb ˜
Á 2j˜
Áb ˜
Á 3j ˜
with the jth column of B, namely Á . ˜ = (b1j b2j…bnj)¢
Á ˜
Á . ˜
Á . ˜
Á ˜
Ë bnj ¯
where A¢ denotes the transpose of matrix A.
Properties of Matrix Multiplication
If A = (aij)m ¥ n, B = (bij)n ¥ p and C = (cij)p ¥ q then
1.
2.
3.
4.
5.
(AB)C = A(BC)
[Associative law]
AIn = ImA = A
AB may not be equal to BA
k(AB) = (kA)B = A(kB) where k is a scalar
If A is a square matrix, then
Am An = Am + n " m, n ΠN
(Am)n = Amn " m, n ΠN.
6. If A is an invertible matrix then
(A–1 B A)m = A–1 B m A
and A–m = (A–1)m " m Œ N.
TRANSPOSE OF A MATRIX
Definition: Let A = (aij)m ¥ n be a matrix. The transpose of
A, denoted by A¢ or by At is the matrix A¢ = (bij)n ¥ m where
bij = aji " i and j.
By A we mean a matrix B = (bij)m ¥ n where bij = a ij where
a denotes the conjugate of a. By A* we mean
A* = ( A )¢ = ( A ¢ )
Properties of Transpose of Matrix
1. (A + B)¢ = A¢ + B¢
2. (kA)¢
= kA¢ where k is a scalar
3. (AB)¢
= B¢A¢
[Reversal law]
4. If A is an invertible matrix, then (A–1)¢ = (A¢)–1.
ADJOINT AND INVERSE OF A MATRIX
Let A = (aij)n ¥ n be square matrix. The adjoint of A is defined
to be the matrix adj A = (bij)n ¥ n where bij = Aji
where Aji is the cofactor of (j, i)th element of A.
Properties of Adjoint
1. A(adj A) = (adj A)A = |A|In
2. adj(kA) = kn–1 (adj A)
3. adj(AB) = (adj B) (adj A)
Definition: A square matrix A is said to be singular if
|A| = 0 and non-singular if |A| π 0.
Definition: Inverse of a square matrix A = (aij)n ¥ n is the
matrix B = (bij)n ¥ n such that
AB = BA = In.
1
In fact
A–1 =
(adj A)
if |A| π 0.
A
Properties of Inverse
Let A and B be two invertible matrices of the same size.
1. Inverse of a matrix if it exists is unique.
2. AA–1 = A–1A = In
3. (A–1)–1 = A
4. (kA)–1 = k–1 A–1 if k π 0
5. (AB)–1 = B –1 A–1
[Reversal law]
Ê a bˆ
Ê d - bˆ
6. For a matrix A = Á
, adj A = Á
Ë c d ˜¯
Ë - c a ˜¯
1
Ê d - bˆ if ad – bc π 0
ad - bc ÁË - c a ˜¯
7. If A is a triangular matrix, then A–1, if it exists is
a triangular matrix of the same kind.
0ˆ
Ê a11 0
Á
In fact if A = a21 a22 0 ˜ and a11a22a33 π 0,
˜
Á
Ë a31 a32 a33 ¯
then
0
0 ˆ
Ê a22 a33
1
–1
Á
- a21a33 a33 a11
A =
0 ˜
˜
a11a22 a33 Á
Ë A13
- a32 a11 a11a22 ¯
and A–1 =
where A13 = cofactor of (1, 3)th element in A i.e.
a21 a22
A13 =
a31 a32
8. If A = diag (l1, l2, …, ln) then A–1 exists if and
only if li π 0 " i and
A–1 = diag (l1–1, l2–1, …, ln–1)
Also, Am= diag (l1m, l2m…lmn) if m Œ N
9. If a square matrix A satisfies the equation
a0 + a1x + a2 x2 + … + ar xr = 0, then A is invertible
if a0 π 0 and its inverse is given by
Matrices 5.3
A–1= –
1
[a1 I + a2 A + … + ar Ar–1]
a0
10. If AB = I, then BA = I and B = A–1
11. If AB = CA = I, then B = C.
SOME DEFINITIONS AND RESULTS
A square matrix A is said to be
1. Symmetric if A¢ = A
2. Skew-symmetric if A¢= – A
3. Hermitian if A* = A
4. Skew Hermitian if A* = – A
5. Orthogonal if AA¢ = A¢A = I or if A¢ = A–1
6. Unitary if AAq = AqA = I
Illustration
1
5 2ˆ
Ê 0
2. A = Á -5 0 7˜ is a skew symmetric matrix
Á
˜
Ë -2 -7 0¯
2 - iˆ
Ê 3
is a Hermitian matrix
3. A = Á
Ë2 + i
7 ˜¯
6 - 5iˆ
Ê 2i
is a skew Hermitian matrix
4. A = Á
Ë -6 - 5i
4i ˜¯
6. A =
1
1
(A + A¢ ) and Q =
(A – A¢ ).
2
2
4. For every square matrix A, matrices A + A¢, AA¢ and
A¢A are symmetric and matrices A – A¢ and A¢ – A
are skew-symmetric.
Results 1. If A is an orthogonal matrix, then |A| π 0. In fact
|A| = ± 1 and A , A¢, A* are also orthogonal.
2. If A and B are two orthogonal matrices, then AB
and BA are both orthogonal matrices.
A square matrix A is said to be
1 nilpotent if Am = O for some m ΠN.
2. idempotent if A2 = A
3. involutory if A2 = I.
P=
Illustration
Ê 2 5 -1ˆ
1. A = Á 5 7 -2˜ is a symmetric matrix
Á
˜
Ë -1 -2 0 ¯
Ê cos q
5. A = Á
Ë sin q
In fact
- sin q ˆ
is an orthogonal matrix
cos q ˜¯
1 Ê 2i 1 ˆ
Á
˜ is a unitary matrix.
5 Ë 1 2i ¯
TIP
If A is a skew symmetric matrix, the I + A and I – A are nonsingular matrices.
Some Results
1. The main diagonal elements of a skew-symmetric
matrix are zeros, i.e. aii = 0 " i.
2. Determinant of a skew-symmetric matrix of odd
order is zero and determinant of a skew-symmetric
matrix of even order is a perfect square.
3. Every square matrix can be written uniquely as a
sum of symmetric and a skew-symmetric matrix, i.e.
if A is a square matrix, then there exists a symmetric
matrix P and a skew-symmetric matrix Q such that
A=P+Q
2
Ê 1 -1ˆ
is a nilpotent matrix as A2 = 0
1. A = Á
Ë 1 -1˜¯
Ê 1 -1 1ˆ
2. A = Á 1 -1 1˜ is an idempotent matrix as A2 = A
Á
˜
Ë 1 -1 1¯
Ê1 / 2
3. A = Á
Ë1 / 2
1/ 2 ˆ
2
˜ is an involutory as A = I.
¯
-1 / 2
Remark
The least value of m for which Am = O, is called the index of
the nilpotent matrix.
Definition: If A = (aij) n × n is a square matrix, the trace
of A denoted by tr (A) is sum of all the main diagonal elements, i.e.
tr (A) = a11 + a22 + … + ann
Properties of Trace
Let A, B be two n × n matrices, and k Œ R.
1. tr (kA) = k tr (A)
2. tr (A + B) = tr (A) + tr (B)
3. tr (AB) = tr (BA)
RANK OF A MATRIX
If A = (aij)m ¥ n is a matrix, and B is its submatrix of order
r, then |B|, the determinant is called as r-rowed minor of A.
Definition Let A = (aij)m ¥ n be a matrix. A positive integer
r is said to be rank of A if
(i) A possesses at least one r-rowed minor which is
different from zero; and
(ii) every (r + 1) rowed minor of A is zero.
5.4
Complete Mathematics—JEE Main
From (ii), it automatically follows that all minors
of higher order are zeros.
We denote rank of A by r(A)
Result The rank of a matrix does not change when the following elementary row operations are applied to the matrix.
(a) Two rows are interchanged (Ri ´ Rj)
(b) A row is multiplied by a non-zero constant,
(Ri Æ kRi, with k π 0)
(c) a constant multiple of another row is added to a
given row (Ri Æ Ri + kRj) where i π j
Note : The arrow Æ means “replaced by”.
Note that the application of these elementary row operations
does not change a singular matrix to a non-singular matrix
or a non-singular matrix to a singular matrix. Therefore,
the order of the largest non-singular square submatrix is
not affected by application of any of the elementary row
operations. Thus, the rank of a matrix does not change by
application of any of the elementary row operations.
A matrix obtained from a given matrix by applying any of
the elementary row operations is said to be equivalent to
it. If A and B are two equivalent matrices, we write A ~ B.
Note that if A ~ B, then r(A) = r (B).
By using the elementary row operations, we shall try
to transform the given matrix in the following form
Ê 1 * * *ˆ
Á 0 1 * *˜
˜
Á
Á 0 0 1 *˜
Á. .
.
.˜
˜
Á
.
.˜
Á. .
˜
Á
.
.˜
Á. .
ÁË 0 0 0 *˜¯
where * stands for zero or non-zero element. That is, we shall
try to make aii as 1 and all the elements below aii as zero.
We illustrate the above procedure by the following
illustration.
Illustration
3
Find the rank of the matrix
Ê 2 -3 1ˆ
A = Á 3 5 7˜
Á
˜
Ë 5 3 8¯
Solution
Step 1: As a first step we must get a 1 in the first column
of A. For this we subtract Row 1 from the Row 2.
Ê 2 -3 1ˆ
Á 3 5 7˜ [R2 Æ R2 – R1]
Á
˜
Ë 5 3 8¯
Ê 2 -3 1ˆ
Á 1 8 6˜
Á
˜
Ë 5 3 8¯
Step 2: We must get a 1 in the upper left corner. For this
we interchange Row 1 and Row 2.
Ê 2 -3 1ˆ
Ê 1 8 6ˆ
Á 1 8 6˜ [R1 ´ R2] Á 2 -3 1˜
Á
Á
˜
˜
Ë 5 3 8¯
Ë 5 3 8¯
Step 3: We must get zeros at the two remaining two places
in the first column. For this we multiply R1 by – 2 and add
it to R2 and multiply R1 by – 5 and add it to R3.
Ê 1 8 6ˆ
Á 2 -3 1˜ [R2 Æ R2 – 2R1,
Á
˜
Ë 5 3 8¯
6 ˆ
Ê1 8
Á
R3 Æ R3 – 5R1] 0 -19 -11˜
Á
˜
Ë 0 -37 -22¯
Step 4: We must have 1 in the second column. This 1
should not be in the first row. Also, you should not be tempted to use R1 to obtain this 1. For, if we try to use R1, then
two zeros obtained in the first column will be destroyed. We
multiply R2 by – 2 and add it to R3.
6 ˆ
6 ˆ
Ê1 8
Ê1 8
Á 0 -19 -11˜ [R3 Æ R3 – 2R1] Á 0 -19 -11˜
Á
Á
˜
˜
Ë 0 -37 -22¯
Ë0 1
0 ¯
Step 5: We now obtain a 1 at the (2, 2)th place. For this we
interchange R2 and R3.
6 ˆ
Ê1 8
Á 0 -19 -11˜ [R2 ´ R3]
Á
˜
Ë0 1
0 ¯
6 ˆ
Ê1 8
Á0 1
0 ˜
Á
˜
Ë 0 -19 -11¯
Step 6: We must get a zero at (3, 2)th place. For this we
multiply R2 by 19 and add it to R3.
6 ˆ
Ê1 8
Á0 1
0 ˜ [R3 Æ R3 + 19R2]
Á
˜
Ë 0 -19 -11¯
Ê1 8 6 ˆ
Á0 1 0 ˜
Á
˜
Ë 0 0 -11¯
This matrix is in the desired triangular form. Recall if A ~ B
then r(A) = r(B). Thus, rank of the given matrix A is equal
to the rank of the matrix
1 8 6
Ê1 8 6 ˆ
Á
˜
B = 0 1 0 . Since |B| = 0 1 0
Á
˜
Ë 0 0 -11¯
0 0 -11
= –11 π 0, r(B) = 3.
Hence, r(A) = 3.
Matrices 5.5
Some Tips
After obtaining 1 at (1, 1)th place and zeros at the remaining
places in the first column, forget the first row. Do not use the
first row to manipulate elements in the second or any other
column. If you try to do so the zeros in the first column will
be destroyed.
After obtaining 1 at (2, 2)th place and zeros at the remaining
places in the second column, forget the second row. Do not
use it for manipulating elements in the remaining columns.
The same remark applies to the remaining columns.
If A is a non-singular matrix of order n ¥ n, then r (A) = n.
SYSTEM OF LINEAR EQUATIONS
Let us consider the following m linear equations in n unknowns:
a11 x1 + a12 x2 + ... + a1n xn = b1
a21 x1 + a22 x2 + ... + a2n xn = b2 …(1)
.
.
.
am1 x1 + am2 x2 + ... + amn xn = bm
where b1, b2, ... bm are not all zero.
Ê a11
Áa
Á 21
Á .
The m ¥ n matrix Á
.
Á
Á .
Á
Ë am1
a12 … a1n ˆ
a22 … a2 n ˜
˜
.
. ˜
is called the
.
.
. ˜
˜
.
. ˜
˜
am 2 … amn ¯
coefficient matrix of the system of linear equations. Using
it, we can now write these equations as follows:
Ê a11
Áa
Á 21
Á .
Á .
Á
Á .
Á
Ë am1
a12 … a1n ˆ Ê x1 ˆ Ê b1 ˆ
a22 … a2 n ˜ Á x2 ˜ Á b2 ˜
˜ Á ˜ Á ˜
.
. ˜ Á .˜ Á . ˜
=
.
.
. ˜ Á .˜ Á . ˜
˜ Á ˜ Á ˜
.
. ˜ Á .˜ Á . ˜
˜ Á ˜ Á ˜
am 2 … amn ¯ Ë xn ¯ Ë bm ¯
We can abbreviate the
where
Ê a11
Áa
Á 21
Á .
A=Á
.
Á
Á .
Á
Ë am1
above matrix equation as AX = B,
a12 … a1n ˆ
a22 … a2 n ˜
˜
.
. ˜
,
.
.
. ˜
˜
.
. ˜
˜
am 2 … amn ¯
Ê
Á
Á
Á
X=Á
Á
Á
Á
Ë
x1 ˆ
x2 ˜
˜
. ˜
and B =
. ˜
˜
. ˜
˜
xn ¯
Ê
Á
Á
Á
Á
Á
Á
Á
Ë
b1 ˆ
b2 ˜
˜
. ˜
. ˜
˜
. ˜
˜
bn ¯
By a solution of (1) we mean a set of values x1, x2, ..., xn
such that (1) reduces to an identity.
The augmented matrix for system of equations
AX = B
is the matrix (A | B). This matrix is obtained by adding
(n + 1)th column to A. The elements of this column are the
constants b1, …, bm.
Result The system of equations
a11 x1 + a12 x2 + … + a1n xn = b1
a21 x1 + a22 x2 + … + a2n xn = b2
…………………………………
…(1)
am1 x1 + am2 x2 + … + amn xn = bm
is consistent (that is, possesses a solution) if and only if the
coefficient matrix
Ê
Á
A=Á
Á
ÁË
a11
a21
…
am1
a12
a22
…
am 2
… a1n ˆ
… a2 n ˜
˜
… … ˜
… amn ˜¯
and the augmented matrix
Ê
Á
(A | B) = Á
Á
ÁË
a11
a21
…
am1
a12
a22
…
am 2
… a1n
… a2 n
… …
… amn
b1 ˆ
b2 ˜
˜
…˜
bm ˜¯
have the same rank.
We split the remaining result in two cases.
Case 1. If the system of equations in (1) is consistent and
m ≥ n, then
(i) if r(A) = r(A | B) = n, then the system of equations has a
unique solution and (m – n) equations are superfluous in the
sense that any solution that satisfies other n equations will
satisfy these (m – n) equations.
(ii) if r(A) = r(A | B) = r < n, then the (n – r) unknowns are
assigned arbitrary values and the remaining r unknowns can
be found in terms of those (n – r) unknowns which have
already been assigned values.
Case 2. If the system of equations in (1) is consistent and
m < n, then
5.6
Complete Mathematics—JEE Main
(i) if r(A) = r(A | B) = m, then (n – m) unknowns can be
assigned arbitrary values and the values of the remaining m unknowns can be found in terms of those (n – m)
unknowns which have already been assigned values. Also,
(m – r) equations are superfluous.
Illustration
4
Consider the system of equations
x + y + z =1
2x + 2y + 2z = 2
3x + 3y + 3z = k
Here
Ê1 ˆ
Ê 1 1 1ˆ
Á
˜
A = 2 2 2 , B = Á 2˜ , and
Á ˜
Á
˜
Ë 3 3 3¯
Ë k¯
(ii) if r(A) = r(A | B) = r < m then (n – r) unknowns can be
assigned arbitrary values and the values of the remaining r
unknowns can be found in terms of those (n – r) unknowns
which have already been assigned values. Also, (m – r)
equations are superfluous.
Finding Inverse by Elementary Row Operations
Ê 0 0 0ˆ
adj A = Á 0 0 0˜
Á
˜
Ë 0 0 0¯
Note that
(adj A) B = O
Also, note that the system of equations has
(1) infinite number of solutions if k = 3
(2) no solution if k π 3.
To find inverse of a square matrix A we begin with the
augmented matrix [A | In]. If a sequence of elementary row
operations transforms this matrix to [In | B], then B is A–1.
However, if at any step we obtain all zeros in a row on the
left of the vertical line, the matrix A is not invertible.
SOLUTION OF A SYSTEM OF EQUATION
AX = B WHEN A IS A SQUARE MATRIX
Remark
Unique Solution
Note this case carefully. Mostly students commit mistake in
this case.
The system of equations
AX = B
has a unique solution if |A| π 0 and it is given by X = A–1 B
SOLUTION OF A SYSTEM OF HOMOGENEOUS LINEAR EQUATIONS AX = 0
No Solution
If |A| = 0 and (Adj A)B π O, the system of equations has no
solution.
The system
AX = 0
has a unique solution if |A| π 0 and it is the trivial solution viz.
x1 = x2 = … = xn = 0.
If |A| = 0, the system has infinite number of solutions.
Also if AX = 0 has at least one non-zero solution, then | A| = 0.
The following tree diagram is helpful.
Infinite Number of Solution or no Solution
If |A| = 0, and (Adj A)B = O, the system of equations has
infinite number of solutions or no solution.
AX = B
B πO
B=O
|A| π 0
|A| = 0
r(A) = r (A B)
r (A) = r (A B)
Trivial
solution
Infinite
number of
solutions
System of
equations is
consistent
System of
equations is
inconsistent
Matrices 5.7
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Èa b ˘
Example 1: The number of 2 × 2 matrices A = Í
˙
Îc d ˚
for which
-1
È1 / a 1 / b ˘
Èa b ˘
Í c d ˙ = Í1 / c 1 / d ˙ , (a, b, c, d Œ R) is
Î
˚
Î
˚
(a) 0
(c) 2
Ans. (a)
(b) 1
(d) infinite
Solution: If ad – bc π 0, then
A–1 =
1 È d -b˘
ad - bc ÍÎ-c a ˙˚
d
1
=
¤ ad = ad – bc
ad - bc
a
¤ bc = 0 ¤ b = 0 or c = 0
Thus,
Therefore,
-1
È1 / a 1 / b ˘
Èa b ˘
Í c d ˙ = Í1 / c 1 / d ˙
Î
˚
Î
˚
is never possible.
TIP
If n > 1, then a n × n matrix cannot have inverse in the above
form.
Èa b ˘
Example 2: Let A = Í
˙ , a, b, c, d Œ R
Îc d ˚
If A5 = A3 + I, then A is
(a) a symmetric matrix
(b) a skew symmetric matrix
(c) an invertible matrix
(d) none of these
Ans. (c)
Solution: A5 = A3 + I
fi A(A4 – A2) = (A4 – A2) A = I
fi A is invertible and A–1 = A4 – A2.
Example 3: Let A and B be two 3 × 3 invertible matrices.
If A + B = AB, then
(a) A–1 + B –1 = O
(b) A–1 + B –1 = B –1 A–1
(c) I – A–1 is invertible
(d) B –1 + I is invertible
Ans. (c)
Solution: A + B = AB fi A = AB – B = (A – I)B.
Premultiplying by A–1 we get
I = A–1A = (I – A–1)B
–1
fi I – A is invertible.
Example 4: Let A be a 3 × 3 matrix and
ÏÊ xˆ
¸
ÔÁ ˜
Ô
S = Ì y x, y, z Œ R ˝
ÔÁË z ˜¯
Ô
Ó
˛
Define f : S Æ S by
Ê xˆ
Ê xˆ
Á
˜
f y = A Á y˜
Á ˜
Á ˜
Ë z¯
Ë z¯
Ê x ˆ Ê 0ˆ
Suppose f Á y˜ = Á 0˜ fi x = y = z = 0.
Á ˜ Á ˜
Ë z ¯ Ë 0¯
Then
(a) f is one-to-one
(b) f cannot be onto
(c) A is not invertible (d) A = O
Ans. (a)
Ê x1 ˆ
Ê x2 ˆ
Á
˜
Solution: Let y1 and Á y2 ˜ be such
Á ˜
Á ˜
Ë z1 ¯
Ë z2 ¯
that
Ê x1 ˆ
Ê x2 ˆ
Á
˜
f y1 = f Á y2 ˜
Á ˜
Á ˜
Ë z1 ¯
Ë z2 ¯
fi
Ê x1 ˆ
Ê x2 ˆ
A Á y1 ˜ = A Á y2 ˜
Á ˜
Á ˜
Ë z1 ¯
Ë z2 ¯
TIP
If P(x) is a polynomial such that P(0) π 0 and P(A) = O, then
A is invertible.
fi
Ê x1 - x2 ˆ Ê 0ˆ
A Á y1 - y2 ˜ = Á 0˜
Á
˜ Á ˜
Ë z1 - z2 ¯ Ë 0¯
Complete Mathematics—JEE Main
5.8
Ê x1 - x2 ˆ Ê 0ˆ
f Á y1 - y2 ˜ = Á 0˜
Á
˜ Á ˜
Ë z1 - z2 ¯ Ë 0¯
x1 – x2 = 0, y1 – y2 = 0, z1 – z2 = 0
x 1 = x 2, y 1 = y 2, z 1 = z 2
fi
fi
fi
Ê x1 ˆ
Áy ˜ =
or
Á 1˜
Ë z1 ¯
Thus, f is one-to-one.
Ê x2 ˆ
Áy ˜
Á 2˜
Ë z2 ¯
Ê a bˆ
, where a, b, c, d ΠR.
Example 5: Let A = Á
Ë c d ˜¯
Example 8: For 1 £ i, j £ 3, let
p /2
and let
(a)
(b)
(c)
(d)
Ans. (b)
A = (aij)3×3. Then
A is a singular matrix
AX = B has a unique solution for every 3×3 matrix B
A is a skew-symmetric matrix.
A2 = I
Solution: For 1 £ i £ 3,
p /2
cos(ix )cos(ix ) dx
0
[integrand is an even function]
p /2
Ê a bˆ
Example 6: Let A = Á
, where a, b, c, d ΠR.
Ë c d ˜¯
=
Ú
[1 + cos(2ix )] dx
0
p /2
sin(2ix ) ˆ ˘
Ê
= Ëx+
2i ¯ ˚˙0
For 1 £ i, j £ 3, i π j,
Then
=
p
p
+0=
2
2
p /2
2
2
2
2
det (A) £ a + b c + d
det (A) £ (a + b) (c + d)
det (A) £ ac + bd
det (A) £ (| a | – | b |) (| c | – | d |)
aij = 2
Solution: det(A) = ad – bc £ |ad – bc|
£ | a || d | + | b || c |
Now,
(a2 + b2) (c2 + d2) – (| a || d | + | b || c |)2
=
=
| a || d | + | b || c | £
a +b
2
c +d
2
Example 7: Let A be a 3×3 matrix such that det(A) = –2.
Then det (–2 A–1) is equal to
(a) 4
(b) – 4
(c) 8
(d) – 2
Ans. (a)
Solution: As A–1 is a 3 × 3 matrix,
det (–2 A–1) = (–2)3 det (A–1)
= (–2)3 (det(A))–1
1
= (-8) Ê- ˆ = 4
Ë 2¯
Ú
[cos((i + j ) x ) + cos((i - j ) x )] dx
0
\
Thus,
sin((i + j )p / 2) sin((i - j )p / 2)
+
i+ j
i- j
1
2
a12 = - + 1 = = a21
3
3
a13 = 0 = a31, and
a23 =
= (| ac | – | bd |)2 ≥ 0
2
cos(ix )cos( jx ) dx
0
– (a2d2 + b2c2 + 2 | a || d | | b || c |)
2
Ú
p /2
= a 2c 2 + b 2c 2 + a 2d 2 + b 2d 2
fi
Ú
aii = 2
(b) det(A) ≥ k2
(d) det(A) £ k
Solution: det(A) = ad – bc £ |ad – bc|
fi
det (A) £ | a || d | + | b || c | £ 2k2
(a)
(b)
(c)
(d)
Ans. (a)
cos(ix )cos( jx ) dx
-p / 2
If a , b , c , d £ k, where k > 0, then
(a) det(A) ≥ 2k2
(c) det(A) £ 2k2
Ans. (c)
Ú
aij =
6
1
+ 1 = = a32
5
5
Ê 1 2/3 0 ˆ
A = Á 2 / 3 1 6 / 5˜
Á
˜
Ë 0 6/5 1 ¯
2
Using C2 Æ C2 - C1 , we get
3
1
0
0
det (A) = 2 / 3 5 / 9 6 / 5
0 6/5 1
5 36
199
=π0
9 25
225
As det (A) π 0, A–1 exists. Therefore, AX = B has a unique
solution viz. X = A–1B.
=
Matrices 5.9
Example 9: The number of values of l for which there
exist a non-zero 3×3 matrix A such that A¢ = lA is:
(a) 0
(b) 1
(c) 2
(d) infinite
Ans. (c)
Solution: We have
A = (A¢)¢ = (lA)¢ = lA¢ = l2A.
As A π 0, l2 = 1 fi l = ±1.
For l = 1 all non-zero symmetric matrices can work as A.
For l = –1, all non-zero skew symmetric matrices can work
as A.
È1 a˘
Example 10: Let A = Í
˙ , where a > 0. Sum of the
Î0 1˚
1
1
1
series S = trace (A) + trace Ê Aˆ + trace ÊÁ A2 ˆ˜ + trace ÊÁ A3 ˆ˜
Ë2 ¯
Ë 23 ¯
Ë 22 ¯
+ … is
(a) 3
(c) 6
Ans. (b)
(b) 4
(d) 8
È0 a ˘
Solution: A = I + B, where B = Í
˙
Î0 0 ˚
As
B2 = O, we get Br = O
" r ≥ 2.
Ar = (I + B)r = I + rB
Thus,
È1 ra ˘
=Í
˙
Î0 1 ˚
" r ≥ 1.
1
1
ˆ
Ê 1
trace Á r Ar ˜ = r (2) = r -1
¯
Ë2
2
2
fi
\
S = 2 +1+
1
2
2
+
" r ≥ 1.
1
2
3
+ =
2
1-
1
2
=4
LEVEL 1
Straight Objective Type Questions
2
È1 4 ˘ È x y ˘
Example 11: If Í
=
Í
˙ , y < 0 then x– y + z
˙
Î2 0 ˚ Î z 0 ˚
is equal to
(a) 5
(b) 2
(c) 1
(d) – 3
Ans. (a)
2
Solution: By the equality of two matrices, x = 1, y = 4,
z=2
fi
x = 1, y = – 2, z = 2 as y < 0.
\
x – y + z =1 + 2 + 2 = 5
È 2˘
Í
Example 12: If A = [1 – 2 3], B = -3˙ , then AB is equal
Í ˙
ÎÍ -1˚˙
to
È 2˘
(a) Í-3˙
Í ˙
ÍÎ -1˙˚
(c) [2 6 – 3]
Ans. (d)
È 2˘
(b) Í 6 ˙
Í ˙
ÍÎ-3˙˚
(d) none of these
È 2˘
Solution: AB = [1 -2 3] Í-3˙ = [2 + 6 - 3] = [5]
Í ˙
ÍÎ -1˙˚
È -i 0 ˘
Example 13: If A = Í
˙ , then A¢ A is equal to
Î0 i˚
(a) I
(b) – iA
(c) – I
(d) iA
Ans. (c)
Solution: We have
È-i 0 ˘ È-i 0 ˘ È-1 0 ˘
A¢ A = Í
˙=Í
˙ =–I
˙Í
Î 0 i ˚ Î 0 i ˚ Î 0 -1˚
È cos a sin a ˘
Example 14: If Aa = Í
˙ , then Aa Ab is
Î- sin a cos a ˚
equal to
(b) Aab
(a) Aa + b
(c) Aa - b
(d) none of these
Ans. (a)
Solution: We have
È cos a sin a ˘ È cos b sin b ˘
Aa Ab = Í
˙
˙Í
Î- sin a cos a ˚ Î- sin b cos b ˚
È cos a cos b - sin a sin b
= Í
Î- sin a cos b - cos a sin b
cos a sin b + sin a cos b ˘
- sin a sin b + cos a cos b ˙˚
È cos (a + b ) sin (a + b ) ˘
= Í
˙ = Aa + b
Î- sin (a + b ) cos (a + b )˚
Example 15: Let A and B be two 2 × 2 matrices.
Consider the statements
(i) AB = O fi A = O or B = O
(ii) AB = I2 fi A = B –1
(iii) (A + B)2 = A2 + 2AB + B2
5.10
Complete Mathematics—JEE Main
Then
(a) (i) is false, (ii) and (iii) are true
(b) (i) and (iii) are false, (ii) is true
(c) (i) and (ii) are false, (iii) is true
(d) (ii) and (iii) are false, (i) is true
Ans. (b)
Solution: (i) is false.
È0 1˘
È1 1 ˘
and B = Í
If
A= Í
˙
˙ , then
Î0 -1˚
Î0 0 ˚
È0 0 ˘
AB = Í
˙=O
Î0 0 ˚
Thus,
AB = O fi
/ A = O or B = O
(iii) is false since matrix multiplication is not commutative.
(ii) is true as product AB is an identity matrix, if and
only if B is inverse of the matrix A.
È-2 5˘
È1 5˘
Example 16: If A – 2B = Í
and 2A – 3B = Í
˙,
˙
Î 0 7˚
Î3 7˚
then matrix B is equal to
È 0 6˘
È- 4 -5˘
(b) Í
(a) Í
˙
˙
Î-3 7˚
Î - 6 -7˚
È2 -1˘
(c) Í
˙
Î3 2˚
Ans. (a)
È6 -1˘
(d) Í
˙
Î0 1˚
Solution: We have
È - 4 - 5˘
B = (2 A - 3 B ) - 2 ( A - 2 B ) = Í
˙
Î - 6 - 7˚
Example 17: If A and B two are 3 × 3 matrices, then
which one of the following is not true:
(a)
(b)
(c)
(d)
Ans. (b)
(A + B) ¢= A ¢ + B ¢
(AB)¢ = A¢ B¢
det (AB) = det (A) det (B)
A (adj A) = |A| I3
Solution: If A and B are two 3 × 3 matrices, then
(AB)¢ = B¢A¢
[Reversal Law]
and not (AB)¢ = A¢ B¢.
Ê cos q - sin q ˆ
Example 18: If A = ÁË
, then
sin q cos q ˜¯
(a) A is an orthogonal matrix
(b) A is a symmetric matrix
(c) A is a skew-symmetric matrix
(d) none of these
Ans. (a)
Solution: We have
Ê cos q
AA¢ = Á
Ë sin q
- sin q ˆ Ê cos q
cos q ˜¯ ÁË - sin q
sin q ˆ
cos q ˜¯
Ê
cos2 q + sin 2 q
= Á
Ë cos q sin q - sin q cos q
cos q sin q - sin q cos q ˆ
˜
¯
sin 2 q + cos2 q
1 0ˆ
= ÊÁ
= I2
Ë 0 1˜¯
Similarly,
A¢ A = I2
Thus, A is an orthogonal matrix.
Example 19: If
Èa 2 ab ac ˘
c -b˘
È0
Í
˙
2
Í
A = Íab b bc ˙ and B = -c 0
a ˙ then the product
Í
˙
Í
2˙
Í
˙˚
0
b
a
Î
ac
bc
c
ÍÎ
˙˚
AB is equal to
(a) O
(b) A
(c) B
(d) I
Ans. (a)
Solution: We have
È0 - abc + abc a 2 c + 0 - a 2 c - a 2 b + a 2 b + 0 ˘
Í
˙
AB = Í 0 - b2 c + b2 c abc + 0 - abc - ab2 + ab2 + 0 ˙ = O
Í
˙
2
2
2
2
ÍÎ 0 - bc + bc ac + 0 - ac - abc + abc + 0 ˙˚
Example 20: If A is an invertible matrix and B is an
orthogonal matrix, of the order same as that of A, then C =
A –1 BA is
(a) an orthogonal matrix
(b) symmetric matrix
(c) skew-symmetric matrix
(d) none of these
Ans. (d)
Ê cos (p / 2) sin (p / 2) ˆ Ê 0 1ˆ
Solution: Let B = Á
=
Ë - sin (p / 2) cos (p / 2)˜¯ ÁË -1 0˜¯
and
Ê 1 3ˆ
A= Á
,
Ë 0 1˜¯
Ê 1 -3ˆ
A-1 = Á
Ë 0 1 ˜¯
Note that B is an orthogonal matrix.
Ê 1 -3ˆ Ê 0 1ˆ Ê 1 3ˆ
C = A–1 BA = Á
Ë 0 1 ˜¯ ÁË -1 0˜¯ ÁË 0 1˜¯
Ê 3 10 ˆ
=Á
Ë -1 -3˜¯
Note that C is neither symmetric, nor skew-symmetric and
nor-orthogonal.
È cos2 a
cos a sin a ˘
Example 21: Let E(a) = Í
˙ . If
ÍÎcos a sin a
sin 2 a ˙˚
a and b differs by an odd multiple of p/2, then E(a) E(b) is a
(a) null matrix
(b) unit matrix
(c) diagonal matrix
(d) orthogonal matrix
Matrices 5.11
Ans. (a)
Solution: We have
E(a) E(b)
È cos2 a
cos b sin b ˘
cos a sin a ˘ È cos2 b
= Í
Í
˙
˙
ÍÎcos a sin a
sin 2 b ˙˚
sin 2 a ˙˚ ÍÎcos b sin b
cos a cos b cos (a - b ) cos a sin b cos (a - b )˘
= È
Í sin a cos b cos (a - b ) sin a sin b cos (a - b ) ˙
Î
˚
As a and b differ by an odd multiple of p/2, a –b =
(2n + 1)p/2 for some integer n. Thus, cos [(2n + 1) p/2] = 0
\
E(a) E(b) = O
2 ˘ È1 0 ˘
È2 1 ˘ È-3
AÍ
Example 22: If Í
˙
˙=Í
˙ , then
Î 7 4 ˚ Î 5 - 3˚ Î 0 1 ˚
matrix A equals
5˘
È 7
È2 1˘
(b) Í
(a) Í
˙
˙
Î-11 - 8˚
Î 5 3˚
È 7 1˘
È 5 3˘
(c) Í
(d) Í
˙
˙
Î34 5˚
Î13 8˚
Ans. (a)
Solution: If XAY = I, then A = X –1 Y –1 = (YX)–1
5˘
È-3 2 ˘ È2 1 ˘ È 8
=Í
In this case
YX = Í
˙
˙
Í
˙
Î 5 -3˚ Î7 4 ˚ Î-11 -7˚
5 ˘ -1 È 7
5˘
È 8
=Í
\
A= Í
˙
˙
Î-11 -8˚
Î-11 -7˚
Example 23: The matrix A satisfying
È1 5˘ È3 -1˘
AÍ
˙ is
˙ =Í
Î0 1˚ Î6 0 ˚
È3 2 ˘
(a) Í
˙
Î6 -3˚
È3 -16 ˘
(b) Í
˙
Î6 -30 ˚
È3 -16 ˘
È3 -3˘
(c) Í
(d) Í
˙
˙
Î6 30 ˚
Î6 2 ˚
Ans. (b)
Solution: We know that if AC = B, then A = BC –1.
\
-1
È3 -1˘ È1 -5˘
È3 -1˘ È1 5˘
= Í
A= Í
˙Í
˙
˙
Í
˙
Î6 0 ˚ Î0 1 ˚
Î6 0 ˚ Î0 1˚
È3 -16 ˘
=Í
˙
Î6 -30 ˚
È1 1 ˘
È3 2 ˘
is the matrix Í
Example 24: If product A Í
˙
˙,
Î2 0 ˚
Î1 1 ˚
–1
then A is given by
È0 -1˘
(a) Í
˙
Î2 - 4˚
1˘
È0
(c) Í
˙
Î2 - 4˚
È 0 -1˘
(b) Í
˙
Î-2 - 4˚
(d) none of these
Ans. (c)
Solution: If AB = C, then B –1 A –1 = C –1
fi
A –1 = BC –1
È1 1 ˘ È3 2 ˘
Here
AÍ
˙
˙ = Í
Î2 0 ˚ Î1 1 ˚
fi
È1
A –1 = Í
Î2
È1
= Í
Î2
1 ˘ È3 2 ˘ -1
0 ˙˚ ÍÎ1 1 ˙˚
1 ˘ È 1 -2 ˘ È0 1 ˘
=
0 ˙˚ ÍÎ-1 3 ˙˚ ÍÎ2 - 4 ˙˚
Example 25: If A and B are two skew-symmetric
matrices of order n, then
(a) AB is a skew-symmetric matrix
(b) AB is a symmetric matrix
(c) AB is a symmetric matrix if A and B commute
(d) none of these
Ans. (c)
Solution: We are given
A¢ = – A and B¢ = –B
Now,
(AB)¢ = B¢ A¢ = (– B) (– A) = BA
= AB if A and B commute.
Example 26: Which of the following statements is false:
(a) If |A| = 0, then |adj A| = 0
(b) Adjoint of a diagonal matrix of order 3 × 3 is a
diagonal matrix
(c) Product of two upper triangular matrices is a
upper triangular matrix
(d) adj (AB) = adj (A) adj (B)
Ans. (d)
Solution: We have
adj (AB) = adj (B) adj (A)
and not
adj (AB) = adj (A) adj (B)
Example 27: If A and B are symmetric matrices, then
AB – BA is a
(a) symmetric matrix
(b) skew-symmetric matrix
(c) diagonal matrix
(d) null matrix
Ans. (b)
Solution: We are given A¢ = A, B¢ = B
Now (AB – BA)¢ = (AB)¢ – (BA)¢ = B¢ A¢ – A¢ B¢
= BA – AB = – (AB – BA)
i.e.
(AB – BA)¢ = – (AB – BA)
Hence, AB – BA is a skew-symmetric matrix.
Example 28: Let A and B be two 3 × 3 matrices, such
that A + B = 2B¢ and 3A + 2B = I, then
(a) A – B = O
(b) A + B = I
(c) A – B = I
(d) A + 2B = O
Ans. (a)
5.12
Complete Mathematics—JEE Main
Solution: A + B = 2B¢
3A + 2B = I
fi
B = 6B¢ – I
fi
B¢ = (6B¢ – I)¢ = 6B – I
Therefore, B = 6(6B – I) – I
fi (1 – 36)B = – 7I
1
fi
B= I
5
2
3
\
3A = I - I = I
5
5
1
fi
A= I
5
\
A =B
Example 29: Let A and B be two non-zero 3 × 3 matrices
such that AB = O. Then
(a) Both A and B are non-singular
(b) Exactly one of A, B is singular
(c) Both A and B are singular
(d) Both A + B and AB are singular
Ans. (c)
Solution: Suppose A is non-singular, then
A–1 (AB) = A–1(O) fi B = O
A contradiction.
\ A is singular
Similarly, B is singular.
Example 30: Suppose A is a 3×3 skew-symmetric
matrix. Let B = (I + A)–1 (I – A). Then
(a) B is orthogonal
(b) B is skew symmetric
(d) B is a diagonal matrix
(c) B2 = O
Ans. (a)
Solution: We have
BB¢ = (I + A)–1 (I – A) [(I + A)–1 (I – A)]¢
= (I + A)–1 (I – A) (I – A)¢ ((I + A)¢]–1
= (I + A)–1 (I – A) (I + A) (I – A)–1
[∵ A¢ = –A]
= (I + A)–1 (I + A) (I – A) (I – A)–1
[I + A and I – A commute]
= (I) (I) = I
Thus, B is orthogonal.
Example 31: Let an = 3n + 5n, n ΠN, and let
Ê an
A = Á an +1
Á
Ë an + 2
an +1
an + 2
an + 3
an + 2 ˆ
an + 3 ˜
˜
an + 4 ¯
Then
(a)
(b)
(c)
(d)
0 is a root of the equation det (A – xI) = 0
det(A) = an an + 2 an + 4
det(A) < 0
det(A) = an + an + 2 + an + 4
Ans. (a)
Solution: Using C2 Æ C2 – 3C1, C3 Æ C3 – 32C1, we get
an
det(A) = an +1
2(5n )
16(5n )
2(5n +1 ) 16(5n +1 )
an + 2 2(5n + 2 ) 16(5n + 2 )
=0
[∵ C2 and C3 are proportional]
Thus, x = 0 is a root of det (A – xI) = 0
Example 32: First row of a matrix A is [1 3 2]. If
È -2 4 a ˘
adj A = Í -1 2 1 ˙
Í
˙
ÎÍ3a -5 -2˚˙
then a possible value of det(A) is
(a) 1
(b) 2
(c) –1
(d) –2
Ans. (a)
Solution: We know that
det(A) = a11 A11 + a12 A12 + a13 A13
= (1) (–2) + (3) (–1) + (2) (3a)
= 6a – 5
and
(det(A))2 = det (adj A) = 11a – 10
Thus,
(6a – 5)2 = 11a – 10
fi
36a2 – 71a + 35 = 0
fi
a = 1, 35/36
Therefore, a possible value of det(A) is 1.
Example 33: Suppose ABC is a triangle with sides a, b,
c and semiperimeter s. Then matrix
s-c
È s
˘
Í
˙ Ès - a ˘
2
2
A = Ís(s - b) (s - a) (s - c)˙ Ís - b ˙
Î
˚ 2 ¥1
Í s( s - c )
(s - a)2 ˙˚3 ¥ 2
Î
bc
È
˘
Í
- ca (s - a )(s - b)˙
Í
˙
ÎÍ ab (s - a) ˚˙3 ¥1
is equal to:
Èa ˘
(a) Íb ˙
Í ˙
ÎÍc ˚˙
Ès ˘
(c) Ís˙
Í ˙
ÍÎs˙˚
Ans. (b)
È0˘
(b) Í0˙
Í ˙
ÎÍ0˚˙
Ès - a ˘
(d) Ís - b ˙
Í
˙
ÍÎs - c ˙˚
Solution:
Ès(s - a) + (s - c)(s - b) - bc
˘
A = Í(s - a )(s - b) (s(s - b) + (s - a )(s - c) - ca)˙
Í
˙
ÍÎ(s - a )(s(s - c) + (s - a )(s - b) - ab)
˙˚
Matrices 5.13
But s(s – a) + (s – c) (s – b) – bc
= s(s – a) + s(s – b – c)
= s(2s – a – b – c) = 0 etc.
È0˘
Thus, A = Í0˙
Í ˙
ÍÎ0˙˚
Example 34: The number of matrices
Èa b ˘
A= Í
˙ (where a, b, c, d Œ R)
Îc d ˚
such that A–1 = – A is :
(a) 0
(b) 1
(c) 2
(d) infinite
Ans. (d)
Solution: As A–1 exists, det(A) = ad – bc π 0.
È- a - b ˘
Also, det (–A) = det Í
˙ = ad – bc
Î -c - d ˚
= det (A)
1
, A–1 = –A implies
since det(A–1) =
det( A)
1
= ad – bc
ad - bc
fi (ad – bc)2 = 1 fi ad – bc = ±1.
If ad – bc = 1, then A–1 = –A
gives
È d - b ˘ È- a - b ˘
A–1 = Í
˙=Í
˙
Î- c a ˚ Î - c - d ˚
fi
a + d = 0.
\ a(–a) – bc = 1
fi
bc = – (1 + a2) fi bc π 0 and
b =–
1 + a2
c
È
(1 + a 2 ) ˘
a
Í
˙
\
A=
c ˙ , where c π 0.
Í
ÍÎc
- a ˙˚
Thus, there are infinite number of such matrices.
TIP
It is unnecessary to consider the case ad – bc = –1
Example 35: Let A be a 3×3 matrix with entries from
the set of real numbers, If the system of equations A2X = 0
has a non-trivial solution, then
(a) AX = 0 has a non-trivial solution.
(b) AX = 0 does not have a non-trivial solution.
(c) A is a non-singular matrix.
(d) none of these.
Ans. (a)
Solution: As A2X = 0 has a non-trivial solution,
det (A2) = 0 fi (det(A))2 = 0
fi
det(A) = 0
\
AX = 0 has a non-trivial solution.
Example 36: Suppose a, b Œ R and a π b.
Èa b ˘
Let
A= Í
˙.
Îb - a ˚
Let M be a matrix such that MA = A2m for some m ΠN, then
M is equal to
(b) (a2 + b2)m–1 A
(a) (a2 + b2)m I
2
2 m–1
(d) (a2 + b2)m A
(c) –(a + b ) A
Ans. (b)
Solution: We have
Èa 2 + b 2
0 ˘
2
2
A2 = Í
˙ = (a + b )I
2
ÍÎ 0
a + b2 ˙˚
fi
A2m = ((a2 + b2) I )m = (a2 + b2)m I
Also,
A–1 =
=
Now,
fi
È- a - b ˘
Í
˙
a + b Î- b a ˚
-1
2
2
Èa b ˘
Í
˙
a + b Îb - a ˚
1
2
2
MA = A2m = (a2 + b2)m I
M = (a2 + b2)m A–1
= (a2 + b2)m–1 A
Èa b ˘
Example 37: Let A = Í
˙ , be a 2 × 2 matrix where
Îc d ˚
a, b, c, d Π{0, 1}. The number of such matrices which have
inverse is
(a) 5
(b) 6
(c) 7
(d) 8
Ans. (b)
Solution: det(A) = ad – bc
Note that det(A) can take value –1, 0 or 1. We have
det(A) = 1 ¤ ad = 1, bc = 0
¤ a = 1, d = 1 or (b = 0, c = 0, b = 0; c = 1;
b = 1, c = 0)
and
det(A) = –1 ¤ ad = 0 or bc = 1
This is also possible in 3 cases.
\ A–1 exists in 6 cases.
Example 38: If D = diag (d1, d2, …., dn) where di π 0,
for i = 1, 2, . . . ., n, then D –1 is equal to
(a) D
(b) 2D
(c) diag (d1–1, d2–1, …., d –1
n)
(d) Adj D
Ans. (c)
Solution: See Theory.
5.14
Complete Mathematics—JEE Main
Example 39: The inverse of a symmetric matrix (if it
exists) is
(a) a symmetric matrix
(b) a skew-symmetric matrix
(c) a diagonal matrix
(d) none of these
Ans. (a)
È1 0 2 ˘
Example 43: If A = Í5 1 x ˙ is a singular matrix,
Í
˙
ÍÎ1 1 1 ˙˚
then x is equal to
(a) 3
(b) 5
(c) 9
(d) 11
Ans. (c)
Solution: Let A be an invertible symmetric matrix.
We have
AA–1 = A–1 A = In
fi
(AA–1)¢ = (A–1 A)¢ = (In)¢
fi
(A–1)¢ A¢ = A¢(A–1)¢ = In
fi
(A–1)¢ A = A(A–1)¢ = In
(A–1)¢ = A–1 [inverse of a matrix is unique]
Example 40: The inverse of a skew-symmetric matrix
(if it exists) is
(a) a symmetric matrix
(b) a skew-symmetric matrix
(c) a diagonal matrix
(d) none of these
Ans. (b)
Solution: We have A¢ = – A
Now,
AA–1 = A–1 A = In
fi
(AA–1)¢ = (A–1 A)¢ = (In)¢
fi
(A–1)¢ A¢ = A¢(A–1)¢ = In
fi (A–1)¢ (–A) = (– A) (A–1)¢ = In
Thus,
(A–1)¢ = – (A–1)
Solution: As A is a singular matrix,
1 0
0
5 1 x - 10 = 0
|A| = 0 fi
1 1
-1
fi
fi
Solution: As A is an orthogonal matrix,
A¢ A = AA¢ = In
fi
|A¢ A| = |AA¢| = |In|
fi
|A¢| |A| = 1 fi |A| |A| = 1
fi
|A|2 = 1 fi |A| = + 1
fi
– 1 – x + 10 = 0
x = 9.
2 ˘
È2 x -1
Í
˙
x 2 x 2 ˙ is singular is
A= Í 1
Í 1 1x 2 ˙
Î
˚
(a) ± 1
(b) ± 2
(c) ± 3
(d) none of these
Ans. (a)
Solution: We have
2
1 x
1 2 x2
Ê 2ˆ x 2x
+
+2
|A| = Á ˜
Ë x¯ 1 x 2
1 1x
1 2
Example 41: The inverse of a skew-symmetric matrix
of odd order is
(a) a symmetric matrix
(b) a skew-symmetric matrix
(c) diagonal matrix
(d) does not exist
Ans. (d)
Example 42: If A is an orthogonal matrix, then |A| is
(a) 1
(b) – 1
(c) ± 1
(d) 0
Ans. (c)
=0
Example 44: The value of x for which the matrix
[inverse of a matrix is unique]
Solution: Let A be a skew-symmetric, matrix of order
n. By definition
A¢ = – A
fi
| A¢| = |– A| fi | A| = (– 1)n |A|
fi
| A| = – |A|
[∵ n is odd]
fi
2|A| = 0 fi |A| = 0
\ A–1 does not exist.
[using C3 Æ C3 – 2C1]
1 x - 10
1
-1
=
Now,
2
Ê1
ˆ
(0 ) + 2 - 2 x 2 + 2 Á - x ˜
Ë
¯
x
x
2
2
2
= 2 x (1 - x ) + 2(1 - x ) = 2( x + 1) (1 - x )
x
x
|A| = 0 fi x = + 1.
Example 45: If square matrix A is such that 3A3 + 2A2 +
5A + I = O, then A–1 is equal to
(a) 3A2 + 2A + 5I
(b) – (3A2 + 2A + 5I)
(c) 3A2 – 2A – 5I
(d) none of these
Ans. (b)
Solution: We have A(3A2 + 2A + 5I)
\
=–I
A–1 = – (3A2 + 2A + 5I). [\ Inverse of a matrix
is unique]
È1 0 ˘
È1 0 ˘
Example 46: If A = Í
and I = Í
˙
˙ then which
Î1 1 ˚
Î0 1 ˚
one of the following holds for all n ≥ 1, by the principle of
mathematical induction.
Matrices 5.15
(a)
(b)
(c)
(d)
Ans. (c)
An = nA + (n – 1) I
An = 2n – 1 A + (n – 1) I
An = nA – (n – 1) I
An = 2n – 1 A – (n – 1) I.
È-1 0 1 ˘ È 2 6 4 ˘ È-3 -5 -5˘
\
A = Í 1 1 3˙ Í 1 0 1 ˙ = Í 0 9 2 ˙
˙
Í
˙Í
˙ Í
ÍÎ 2 0 2 ˙˚ ÍÎ-1 1 -1˙˚ ÎÍ 2 14 6 ˙˚
Example 49: If w is a complex cube root of unity, then
È 1 w2 w ˘
Í
˙
1 ˙ is a
the matrix A = Íw 2 w
Í
˙
1 w 2 ˙˚
ÍÎ w
(a) singular matrix
(b) non-singular matrix
(c) skew-symmetric matrix
(d) none of these
Ans. (a)
–1
È1 0 ˘
Solution: For n = 2, A2 = Í
˙
Î2 1 ˚
È3 0 ˘
2
For n = 2, RHS of (a) = 2A + I = 3 Í
˙ πA
Î2 3 ˚
For n = 2, RHS of (b) = 2A + I π A2
So possible answer is (c) or (d).
È1 0 ˘
In fact An = Í
˙ which equals nA – (n – 1)I.
În 1 ˚
Solution: We have
Alternatively. Write A = I + B
È0 0 ˘
where B = Í
˙
Î1 0 ˚
As B2 = 0, we get Br = 0 " r ≥ 2.
By the binomial theorem
An = I + nB = I + n(A – I) = nA – (n – 1)I.
Example 47: Let A, B, C be three square matrices of the
same order, such that whenever AB = AC then B = C, if A is
(a) singular
(b) non-singular
(c) symmetric
(d) skew-symmetric
Ans. (b)
Solution: If A is non-singular, A–1 exists.
Thus,
AB = AC
fi A–1 (AB) = A–1 (AC)
fi
(A–1 A) B = (A–1 A) C fi IB = IC
fi
B = C.
È 2 6 4˘
Example 48: Let B = Í 1 0 1 ˙ . If the product BA
Í
˙
ÍÎ -1 1 -1 ˙˚
È -3 5 5 ˘
(b) Í 0 0 9 ˙
Í
˙
ÍÎ 2 14 16 ˙˚
È -3 -5 -5 ˘
(c) Í 0 0 2 ˙
Í
˙
ÍÎ 2 14 6 ˙˚
Ans. (c)
È -3 -3 -5 ˘
(d) Í 0 9 2 ˙
Í
˙
ÍÎ 2 14 6 ˙˚
Solution: We have (BA)–1 = C
fi A–1 = CB
fi
A–1 B–1 = C
w2
w
|A| = w 2
w
1 = w2 + w +1
w
2
1
w
1
w
2
1+ w2 + w w2
w +1+w
w
1 =0
w2
using C1 Æ C1 + C2 + C3 and 1 + w + w 2 = 0
\ A is a singular matrix.
È0 1 2 ˘
Example 50: If A = Í1 2 3˙ and
Í
˙
ÎÍ3 x 1 ˚˙
È1/ 2 -1/ 2 1/ 2 ˘
Í
y ˙ , then
A = -4
3
Í
˙
ÍÎ5 / 2 -3 / 2 1/ 2 ˙˚
(a) x = 1, y = – 1
(b) x = – 1, y = 1
(c) x = 2, y = – 1/2
(d) x = 1/2, y = 1/2
Ans. (a)
–1
Solution: We have
È1 0 0 ˘
È0 1 2 ˘ È1/ 2 -1/ 2 1 / 2 ˘
Í0 1 0 ˙ = AA-1 = Í1 2 3˙ Í -4
3
y ˙
Í
˙
Í
˙Í
˙
ÍÎ0 0 1 ˙˚
ÍÎ3 x 1 ˙˚ ÍÎ5 / 2 -3 / 2 1/ 2 ˙˚
0
y +1 ˘
È 1
Í
0
1
2( y + 1)˙
=
Í
˙
ÎÍ4(1 - x ) 3( x - 1) 2 + xy ˙˚
È -1 0 1 ˘
has inverse C = Í 1 1 3 ˙ , then A–1 equals
Í
˙
ÍÎ 2 0 2 ˙˚
È -3 -5 5 ˘
(a) Í 0 9 14 ˙
Í
˙
ÍÎ 2 2 6 ˙˚
1
fi
1 – x = 0, x – 1= 0,
y +1 = 0, y + 1= 0, 2 + xy = 1
\
x = 1, y = – 1.
Èa b ˘
Example 51: Let A = Í
˙ , where a, b, c, d Œ R. If
Îc d ˚
A Рa I is invertible for all a ΠR, then
(a) bc > 0
(b) bc = 0
Ê 1 ˆ
(d) a = 0
(c) bc > min Ë0, ad ¯
2
5.16
Complete Mathematics—JEE Main
Ans. (c)
Solution: As A Рa I is invertible for all a ΠR.
det (A – a I) π 0
" a ΠR.
fi
(a – a) (d – a) – bc π 0
" a ΠR.
fi a – (a + d)a + ad – bc π 0
" a ΠR.
2
Solution: A2 – A + I = 0 fi I = A – A2 = A(I – A).
Thus, A–1 = I – A.
Ê 1 -1 1 ˆ
Example 55: Let A = Á 2 1 -3˜ and
Á
˜
Ë1 1 1 ¯
Ê 4 2 2ˆ
(10)B = Á - 5 0 a ˜
Á
˜
Ë 1 2 3¯
Therefore
(a + d)2 – 4(ad – bc) < 0
fi
(a – d)2 + 4 bc < 0
Therefore, bc < 0.
Also, a2 + d2 – 2ad + 4bc < 0
fi
0 £ a2 + d2 < 2ad – 4bc
1
ad.
2
1
Thus,
bc < min Ê0, ad ˆ
Ë 2 ¯
È1 2 ˘
2
Example 52: If A = Í
˙ , then A – 5A – I equals
3
4
Î
˚
(a) O
(b) I
(c) 2I
(d) none of these
Ans. (b)
fi
If B is the inverse of matrix A, then a equals
(a) 2
(b) –1
(c) –2
(d) 5
Ans. (d)
Solution: 10I = 10(AB) = A(10B)
bc <
È1 2 ˘ È1 2 ˘ È 7 10 ˘
Solution: A = Í
˙Í
˙=Í
˙
Î3 4 ˚ Î3 4 ˚ Î15 22 ˚
We have
È 7 10 ˘ È1 2 ˘ È2 0 ˘
A2 – 5A = Í
˙ -5 Í
˙=Í
˙ = 2I
Î15 22 ˚ Î3 4 ˚ Î0 2 ˚
fi
Ê 10 0 5 - a ˆ
= Á 0 10 a - 5˜
Á
˜
Ë 0 0 5 + a¯
fi a = 5.
2
A2 – 5A – I = I
fi
È3 2 ˘
2
Example 53: Let A = Í
˙ satisfies A + aA + bI = O,
Î1 1 ˚
then a, b are respectively equal to
(a) – 4, 2
(b) –3, 3
(c) – 4, 1
(d) –3, 1
Ans. (c)
È3 2 ˘ È3 2 ˘ È11 8˘
Solution: A2 = Í
˙Í
˙=Í
˙
Î1 1 ˚ Î1 1 ˚ Î 4 3˚
Now
È11 + 3a + b 8 + 2 a ˘
O = A2 + aA + bI = Í
3 + a + b ˙˚
Î 4+a
fi
fi
3a + b = –11, 4 + a = 0, a + b = –3
a = –4, b = 1.
2
Example 54: Let A – A + I = O, then inverse of A is
(a) I – A
(b) A – I
(c) A + I
(d) A
Ans. (b)
Ê 10 0 0 ˆ Ê 1 -1 1 ˆ Ê 4 2 2 ˆ
Á 0 10 0 ˜ = Á 2 1 -3˜ Á -5 0 a ˜
Á
˜ Á
˜Á
˜
Ë 0 0 10¯ Ë 1 1 1 ¯ Ë 1 -2 3 ¯
Èa b ˘
2
Example 56: If A = Í
˙ and A =
b
a
Î
˚
(a)
(b)
(c)
(d)
Ans. (a)
Èa b ˘
Í b a ˙ , then
Î
˚
a = a2 + b2 b = 2ab
a = a 2 + b 2, b = a 2 – b 2
a = 2ab, b = a2 + b2
a = a2 + b2, b = –2ab
2
2
Èa b ˘ Èa b ˘ Èa + b
=
Solution: A2 = Í
˙Í
˙ Í
Îb a ˚ Îb a ˚ ÍÎ 2ab
\
fi
Èa 2 + b2
Èa b ˘
=
Í
Íb a ˙
Î
˚
ÍÎ 2ab
2ab ˘
˙
a 2 + b2 ˙˚
2ab ˘
˙
a + b2 ˙˚
2
a = a2 + b2, b = 2ab
Example 57: Let w π 1 be a cube root of unity and S be
the set of all non-singular matrices of the form
È1
Í
A= Íw
Íw 2
Î
a
1
w
b˘
˙
c˙
1 ˙˚
where a, b, c are either w or w2. Then number of distinct
matrices in the set S is
(a) 2
(b) 6
(c) 4
(d) 8
Ans. (a)
Matrices 5.17
Ans. (d)
Solution: We have
|A| = 1
w
1 c
-a 2
w 1
w
c
1
+b
w
1
w2
w
Solution: Note that A2 = 49I
A–1 = 1 A
49
fi
2
= 1 – cw – aw + acw = (1 – aw)(1 – cw)
A is non-singular ¤ |A| π 0 ¤ a = c = w.
Note that b can take any value w or w2.
Thus, there are two non-singular matrices.
Example 58: If a matrix A is both symmetric and skewsymmetric, then
(a) A is a diagonal matrix
(b) A is a scalar matrix
(c) A is a zero matrix
(d) none of these
Ans. (c)
\
Solution: We are given A¢ = A and A¢ = –A.
A = –A fi 2A = O fi A = O
È2 -1˘
È5 2 ˘
È 2 5˘
Example 59: Let A = Í
,B= Í
,C= Í
˙
˙
˙.
Î3 4 ˚
Î7 4 ˚
Î3 8 ˚
Let D be a matrix such that CD = AB, then D equals
(a) I
(b) O
(c) –A
(d) none of these
Ans. (d)
Solution: CD = AB
Now
D = C –1 (AB)
È2 -1˘ È5 2 ˘ È 3 0 ˘
AB = Í
˙Í
˙=Í
˙ and
Î3 4 ˚ Î7 4 ˚ Î43 22 ˚
C
\
fi
–1
Example 62: Let A = [5 3] and B = [3 7]
The number of non-zero matrices C such that AC = BC is
(a) 0
(b) 1
(c) infinitely many
(d) none of these
Ans. (c)
È2x ˘
Solution: Note that every matrix of the form Í ˙ , x Œ R
Îx˚
satisfies the required condition.
È0 2y z ˘
Í
Example 63: If A = x y - z ˙ satisfies A¢ = A – 1,
Í
˙
then
ÍÎ x - y z ˙˚
(a) x = ± 1/ 6 , y = ± 1/ 6 , z = ± 1/ 3
(b) x = ± 1/ 2 , y = ± 1/ 6 , z = ± 1/ 3
(c) x = ± 1/ 2 , y = ± 1/ 6 , z = ± 1/ 3
(d) x = ± 1/ 2 , y = ± 1/3, z = ± 1/ 2
Ans. (b)
Solution: A¢ = A – 1 ¤ AA¢ = I
Now,
È 8 -5˘
= Í
˙
Î-3 2 ˚
È 8 -5˘ È 3 0 ˘ È-191 -110 ˘
D= Í
˙Í
˙ = Í
44 ˙˚
Î-3 2 ˚ Î43 22 ˚ Î 77
Example 60: If A2 = A, the (I + A)4 equals
(a) I + 15A
(b) I + 7A
(c) I + 8A
(d) I + 11A
Ans. (a)
Solution: As I and A commute, we can apply binomial
theorem to expand (I + A)4. We have
(I + A)4 = I + 4A + 6A2 + 4A3 + A4.
But
A3 = A2 = A, A4 = A3 = A
\
(I + A)4 = I + 4A + 6A + 4A + A = I + 15A
0 - 7˘
È 0
Í
0 ˙ is a
Example 61: The matrix A = 0 - 7
Í
˙
ÍÎ- 7
0
0 ˙˚
(a) scalar matrix
(b) diagonal matrix
(c) unit matrix
(d) non-singular
[\ A is non-singular]
Thus,
fi
x
x˘
È0 2y z ˘ È 0
Í
˙
Í
AA¢ = x y - z
2 y y - y˙
Í
˙ Í
˙
ÍÎ x - y z ˙˚ ÎÍ z - z z ˙˚
È 4 y2 + z 2
Í
= Í 2 y2 - z 2
Í
2
2
ÍÎ- 2 y + z
AA¢ = I
4y2 + z2 = 1,
x2 + y2 + z2 = 1,
\
2 y2 - z 2
x 2 + y2 + z 2
x 2 - y2 - z 2
- 2 y2 + z 2 ˘
˙
x 2 - y2 - z 2 ˙
˙
x 2 + y 2 + z 2 ˙˚
2y2 – z2 = 0
x2 – y2 – z2 = 0
x = ± 1/ 2 ,
y = ± 1/ 6 ,
z = ± 1/ 3
Example 64: Suppose A is square matrix such that
A3 = I, then (A + I)3 + (A – I)3 – 6A equals
(a) I
(b) 2I
(c) A
(d) 3A
Ans. (b)
Solution: Since AI = IA, we have
(A + I)3 + (A – I)3 – 6A
= 2A3 + 2(3AI2) – 6A = 2I + 6A – 6A = 2I
5.18
Complete Mathematics—JEE Main
Example 65: The number of 3 ¥ 3 matrices A whose
entries are either 0 or 1 and for which the system of equations
È2 ˘
Èx˘
Í
˙
A y = Í0 ˙ has exactly five distinct solutions is
Í ˙
Í ˙
ÍÎ z ˙˚
ÍÎ3˙˚
(a) 0
(b) 511
(c) 1024
(d) 5
Ans. (a)
Solution: We know that the system of equation
AX = B
has either (i) unique solution
or
(ii) infinite number of solutions
or
(iii) no solution.
Hence, there cannot exist any matrix A such that
È x ˘ È2 ˘
A Í y ˙ = Í0 ˙
Í ˙ Í ˙
ÍÎ z ˙˚ ÍÎ3˙˚
has five distinct solutions.
Example 66: The number of 3 ¥ 3 non-singular matrices,
with four entries as 1 and all other entries as 0 is
(a) 6
(b) at least 7
(c) less than 4
(d) 5
Ans. (b)
Solution: The matrix
È1 a b ˘
Íc 1 d ˙
Í
˙
ÎÍe f 1 ˚˙
where exactly one of a, b, c, d, e, f is 1 and rest of them are
zeros, is invertible.
There are six such matrices.
È1 0 1 ˘
Also, the matrix Í0 1 0 ˙ is invertible.
Í
˙
ÍÎ1 0 0 ˙˚
Thus, there are at least 7 such matrices which are invertible.
Example 67: Consider the system of linear equations:
x1 +2x2 + x3 = 3
2x1 +3x2 + x3 = 3
3x1 +5x2 + 2x3 = 1
The system has
(a) a unique solution
(b) no solution
(c) infinite number of solutions
(d) exactly 3 solutions
Ans. (b)
Solution: Adding the first two equations and subtracting the third from the sum, we obtain
(x1 + 2x2 + x3) + (2x1 + 3x2 + x3) – (3x1 + 5x2 + 2x3) = 3 + 3 – 1
fi
0= 5
Thus, the system of equations has no solution.
Example 68: Let a, b and c be three real numbers
satisfying
È1 9 7 ˘
[a b c] Í8 2 7˙ = [0 0 0]
Í
˙
ÎÍ7 3 7˙˚
(1)
If the point P(a, b, c) with reference to (1), lies on the plane
2x + y + z = 1, then the value of 7a + b + c is
(a) 0
(b) 12
(c) 7
(d) 6
Ans. (d)
Solution: We have
a + 8b + 7c = 0, 9a + 2b + 3c = 0, 7a + 7b + 7c = 0
Subtracting the last equation from first and multiplying the
last equation by (9/7) and subtracting from the second equation we obtain – 6a + b = 0, 7b + 6c = 0
c
a
b
fi
=
=
= k (say)
–7
1
6
Thus, a = k, b = 6 k, c = – 7 k.
As (a, b, c) = (k, 6 k, – 7 k) lies on 2x + y + z = 1, we get k = 1.
Thus, 7a + b + c = 7k + 6k – 7k = 6k = 6
Example 69: Let P and Q be 3 ¥ 3 matrices with P π Q
If P3 = Q3 and P2 Q = Q2P, then determinant of(P2 + Q2) is
equal to
(a) 1
(b) 0
(c) – 1
(d) –2
Ans. (d)
Solution: P3 = Q3 and P2Q = Q2P gives
P 3 – P 2Q = Q 3– Q 2P
fi
P2(P – Q) = – Q2(P – Q)
fi (P2 + Q2) (P – Q) = 0
If det (P2 + Q2) π 0, then P2 + Q2 is invertible and hence P
= Q. Therefore, det (P2 + Q2) = 0.
Ê1
Example 70: Let A = Á 2
Á
Ë3
0
1
2
Ê1 ˆ
column matrices such that Au1 = Á 0˜
Á ˜
u1 + u2 equals
Ë 0¯
0ˆ
0˜ . If u1 and u2 are
˜
1¯
Ê 0ˆ
and Au2 = Á 1 ˜ , then
Á ˜
Ë 0¯
Ê -1ˆ
(a) Á 1˜
Á ˜
Ë -1¯
Ê -1ˆ
(b) Á -1˜
Á ˜
Ë 0¯
Ê 1ˆ
(c) Á -1˜
Á ˜
Ë -1¯
Ê -1ˆ
(d) Á 1˜
Á ˜
Ë 0¯
Matrices 5.19
Ans. (c)
Ans. (a)
Solution: We have
Solution: The augmented matrix is given by
Ê1 ˆ
Ê 0ˆ
Ê1 ˆ
A(u1 + u2) = Au1 + Au2 = Á 0˜ + Á 1˜ = Á 1 ˜
Á ˜
Á ˜
Á ˜
Ë 0¯
Ë 0¯
Ë 0¯
We solve the above equation for u1 + u2.
We consider the augmented matrix
Ê1
(A|B) = Á 2
Á
Ë3
0
1
2
Ê1
(A|B) ~ Á 5
Á
Ë2
0 1ˆ
0 1˜
˜
1 0¯
0
1
0
01 ˆ
0 -1˜
˜
1 -1¯
fi
(b) PX = X
(d) PX= –X
Solution: We have
P = (P¢)¢ = (2P + I )¢ = 2P¢ + I
P = 2(2P + I ) + I fi P = –I
Thus, there exists X π 0 such that PX = –X
Ê 4 3ˆ
Example 72: Let A = Á
and x, y ΠR are such that
Ë 2 5˜¯
A2 + xA + yI = 0 then (x, y) equals
(a) (–9, –14)
(b) (9, –14)
(c) (–9, 14)
(d) (9, 14)
Ans. (c)
Ê 22 27ˆ
Solution: We have A2 = Á
Ë 18 31 ˜¯
2
\
A + xA + yI = O
fi
22 + 4x + y = 0, 27 + 3x = 0
18 + 2x = 0, 31 + 5x + y = 0
fi
x = –9, y = 14
Example 73: The system of equations
Ê3 - 2 1ˆ
Á 5 - 8 9˜
˜
Á
Ë2
1 a¯
1 bˆ
9 3˜
˜
a -1¯
Ê xˆ
Ê bˆ
Á y˜ = Á 3˜
Á ˜
Á ˜
Ë z¯
Ë -1¯
has no solution if a and b are
(a) a = –3, b π 1/3
(b) a = 2/3, b π 1/3
(c) a π 1/4, b = 1/3
(d) a π –3, b π 1/3
4
-8
1
- 7 2b - 3ˆ
9 3 ˜
˜
a -1 ¯
Applying R2 Æ R2 – 5R1, R3 Æ R3 – 2R1, we get
4
Ê1
Á
(A|B) ~ 0 - 28
Á
Ë 0 -7
Ê 1ˆ
u1 + u2 = Á -1˜
Á ˜
Ë -1¯
Example 71: If P is a 3 ¥ 3 matrix such that
P´ = 2P + I, then there exists a column matrix
Ê xˆ
X = Á y˜ π O such that
Á ˜
Ë z¯
(a) PX = O
(c) PX = 2X
Ans. (d)
-2
-8
1
Applying R1 Æ 2R1 – R2, we get
Applying R3 Æ R3 – 2R2 + R1 and R2 Æ R2 – 2R1, we get
Ê1
(A|B) ~ Á 0
Á
Ë0
Ê3
(A|B) = Á 5
Á
Ë2
-7
2b - 3 ˆ
44 18 - 10b˜
˜
a + 14 5 - 4b ¯
The system of equations will have no solution if
fi
fi
-28
44
18 - 10b
=
π
5 - 4b
-7 a + 14
a + 14 = 11 and 20 – 16b π 18 –10b
a = –3 and b π –1/3
Example 74: Suppose I + A is non-singular. Let B =
(I + A)–1 and C = I – A, then
(a) BC = CB
(b) BC = O
(c) BC = I
(d) none of these
Ans. (a)
Solution: Write C = 2I – (I + A), then
BC = 2(I + A)–1 – I = CB
Ê a bˆ
Example 75: Let A = Á
be such that A3 = O, but
Ë c d ˜¯
A π O, then
(a) A2 = O
(b) A2 = A
2
(d) none of these
(c) A = I – A
Ans. (a)
Solution: As A3 = O, we get |A3| = 0
|A|3 = 0 fi |A| fi ad – bc = 0.
Also,
Ê a 2 + bc (a + d ) bˆ
A2 = Á
˜
Ë (a + d )c bc + d 2 ¯
Ê a 2 + ad (a + d )bˆ
= Á
˜
Ë (a + d )c ad + d 2 ¯
= (a + d)A
If a + d = 0, we get A2 = O.
Suppose a + d π 0, then
A3 = A2A = (a + d)A2
fi
O = (a + d)A2 fi A2 = O
5.20
Complete Mathematics—JEE Main
Assertion-Reason Type Questions
Example 76: Let A be a 2 ¥ 2 matrix.
Statement-1: adj(adj A) = A
Statement-2: |adj A| = |A|
Ans. (b)
Èa b ˘
Solution: Let A = Í
˙ then adj A =
Îc d ˚
fi
È d - b˘
Í- c a ˙
Î
˚
Èa b ˘
=A
d ˙˚
adj (adj A) = Í
Îc
\
Statement-1 is true.
Also, |adj A| = ad – bc = |A|
\ Statement-2 is true
But statement-2 is not a correct explanation for statement-1.
Example 77: Let A be a 2 ¥ 2 matrix with non-zero
entries and let A2 = I, where I is 2 ¥ 2 identity matrix. Define
Tr (A) = sum of diagonal elements of A, and
|A| = determinant of matrix A.
Statement-1: Tr (A) = 0
Statement-2: |A| = 1.
Ans. (c)
Èa b ˘
Solution: Let A = Í
˙
Îc d ˚
Now,
A2 = I fi |A2| = |I |
fi
|A|2 = 1 fi |A| = ± 1.
Suppose |A| = 1.
In this case, A2 = I fi A = A – 1
Ê a bˆ
Ê d - bˆ
fi
ÁË c d ˜¯ = ÁË - c a ˜¯
fi
a = d, b = 0, c = 0
Tr (A) = 0 fi a + d = 0
fi
2a = 0 fi a = 0
Ê 0 0ˆ
In this case A = Á
Ë 0 0˜¯
Thus, if we assume statement-2 is true then we get statement-2 is false.
In case |A| = – 1, then
A = – A–1
Ê a bˆ
Ê d - bˆ
fi
ÁË c d ˜¯ = - ÁË -c a ˜¯ fi a + d = 0
|A| = – 1 fi Tr(A) = 0.
\
Therefore statement-1 is true and statement-2 is false.
Example 78: Let A be a 2 ¥ 2 matrix with real entries.
Let I be the 2 ¥ 2 identity matrix. Denote by Tr (A), the sum
of diagonal entries of A. Assume that A2 = I.
Statement-1: If A π I and A π – I, then det (A) = – 1.
Statement-1: If A π I and A π – I, then Tr (A) π 0.
Ans. (c)
Ê a bˆ
Solution: Let A = Á
Ë c d ˜¯
Now,
A2 = I fi det (A2) = 1
fi (det A)2 = 1 fi det A = ± 1.
Also,
A2 = I fi A = A –1
Ê a bˆ
Ê d – bˆ
fi
ÁË c d ˜¯ = det A ÁË – c a ˜¯
If det A = 1, then
a = d, b = – b, c = – c fi a = d, b = c = 0.
Ê a 0ˆ
.
In this case A = ÁË
0 a˜¯
| A| = 1 fi a2 = 1 fi a = ± 1.
\
A = I or A = – I. A contradiction.
Thus, det (A) = – 1.
Ê a bˆ
Ê d - bˆ Ê - d b ˆ
=
=–1Á
\ Á
˜
Ë c d¯
Ë - c a ˜¯ ÁË c - a˜¯
fi
a = – d fi Tr (A) = a + d = 0.
\ Statement -1 is true and Statement-2 is false.
Example 79: Let
Ê cos q + sin q
A(q) = Á
Ë - 2 sin q
2 sin q ˆ
˜
cos q - sin q ¯
Statement-1: A(p/3)3 = – I
Statement-2: A(q) A(j) = A(q + j)
Ans. (a)
Ê a bˆ
Solution: A(q) A(j) = Á
where
Ë -b c ˜¯
a = (cos q + sin q) (cos j + sin j) – 2 sin q sin j
= (cos q cos j – sin q sin j) + (sin q cos j + cos q sin j)
= cos (q + j) + sin (q + j)
b=
2 [sin j (cos q + sin q) + sin q (sin j – cos f)]
= 2 sin (q + j)
and c = – 2 sin q sin j + (cos q – sin q) (cos j – sin j)
= cos q cos j – sin q sin j – (sin q cos j
+ cos j sin q)
= cos (q + j) – sin (q + j)
Thus, A(q) A(j) = A(q + j)
Matrices 5.21
A(q)2 = A(2q)
A(q)3 = A(2q) A(q) = A(3q).
\
A(p /3)3 = A(p) = – I.
Ê a bˆ
Example 80: Suppose X = Á
satisfies the equation
Ë c d ˜¯
X2 – 4X + 3I = O
Statement-1: If a + d π 4, then there are just two such
matrices X.
Statement-2: There are infinite number of matrices X, satisfying X2 – 4X + 3I = 0.
Ans. (b)
Solution: X2 – 4X + 3I = O
fi
(X – I) (X – 3I) = O
b ( a + d - 4 ) ˆ Ê 0 0ˆ
Ê ( a - 1) ( a - 3) + bc
=
ÁË c ( a + d - 4 )
(d - 1) (d - 3) + bc˜¯ ÁË 0 0¯˜
If a + d π 4, then b = 0, c = 0.
(a – 1) (a – 3) = 0, (d – 1) (d – 3) = 0.
fi
a = 1, 3, d = 1, 3
As a + d π 4, a = 1, d = 1 or a = 3, d = 3.
fi
Ê 1 0ˆ
Ê 3 0ˆ
or X = Á
X= Á
˜
Ë 0 1¯
Ë 0 3˜¯
If a + d = 4, we get
0
ˆ
Ê ( a - 1) ( a - 3) + bc
= O.
ÁË
0
(1 - a ) (3 - a ) + bc˜¯
\ There are infinite number of matrices satisfying X2 –
4X + 3I = 0 as there are infinite number of values of a, b, c
satisfying the relation a2 – 4a + 3 + bc = 0.
Ê 0 2ˆ
Ê1 0 ˆ
Example 81: Suppose A = Á
and B = Á
.
˜
Ë 2 0˜¯
Ë 0 - 1¯
Thus,
Let X be a 2 ¥ 2 matrix such that
X¢AX =B.
Statement-1: X is non-singular and det (X) = ± 2
Statement-2: X is a singular matrix.
Ans. (c)
Solution: If X = O, then X¢AX = O fi B = O.
A contradiction.
Let
det (X) = a, then det (X¢) = a
\
det (X¢AX) = det (B)
fi
a(– 1)a = – 4 fi a = ± 2.
As det (X) π 0, X cannot be a singular matrix.
a12 ˘
È x1 ˘
È y1 ˘
Èa
Example 82: Let A = Í 11
,X= Í ˙,Y= Í ˙
˙
Îa21 a22 ˚
Î x2 ˚
Î y2 ˚
Statement-1: If X¢AX = O for each X, then A must be a
skew-symmetric matrix.
Statement-2: If A is symmetric and X¢AX = O for each X,
then A = O
Ans. (b)
È1 ˘
Solution: Let E1 = Í ˙ , E2 =
Î0 ˚
È0 ˘
Í1 ˙ .
Î ˚
X¢AX = O " X,
E1¢AE1 = O fi a11 = 0
E2¢AE2 = O fi a22 = 0.
As
Next,
fi
(E1 + E2)¢ A(E1 + E2) = O
a12 + a21 = 0
È 0
Thus, A = Í
Î – a12
a12 ˘
is a skew-symmetric matrix.
0 ˙˚
In case A is symmetric, a12 = a21
\
2a12 = 0 fi a12 = 0.
Thus, A = O, in this case.
Example 83: Suppose A, B and C are three 3×3 matrices,
such that AB is invertible.
Statement-1: If AB = AC then B = C.
Statement-2: A is a invertible.
Ans. (a)
fi
fi
fi
fi
Solution: AB is invertible
det(AB) π 0
det(A) det(B) π 0
det(A) π 0
A is invertible
Premultiplying both the sides of
AB = AC
by A–1, we get
A–1 (AB) = A–1 (AC)
fi
(A–1A)B = (A–1A)C
fi
IB = IC or B = C.
Example 84: Suppose a, b, g are real numbers such that
cos(b – g ) + cos(g – a) + cos (a – b ) = –3/2
(1)
Statement-1: The matrix
Ê cos a sin a -2ˆ
A = Á cos b sin b 1 ˜
Á
˜
Ë cos g sin g 1 ¯
is a singular matrix.
Statement-2: cos a + cosb + cos g = 0
and
sin a + sin b + sin g = 0
Ans. (a)
Solution: Write (1) as
cos2 a + cos2 b + cos2 g
+ sin2 a + sin2 b + sin2 g
+ 2cos b cos g + 2 sin b sin g
+ 2cos g cos a + 2 sin g sin a
+ 2cos a cos b + 2 sin a sin b = 0
5.22
Complete Mathematics—JEE Main
(cos a + cos b + cos g )2
fi
Statement-2: If AX = O implies X = O, then A is non-singular.
Ans. (a)
2
+ (sin a + sin b + sin g ) = 0
fi
cos a + cos b + cos g = 0
and sin a + sin b + sin g = 0
Using R1 Æ R1 + R2 + R3, we get R1 consists of all zeros,
therefore det(A) = 0.
Thus, A is a singular matrix.
Example 85: Let A = (aij)3×3 be a 3 × 3 matrix with aij Œ
R for 1 £ i £ 3, 1 £ j £ 3. Let X be a 3 × 1 matrix.
Statement-1: If A is skew-symmetric, then I + A is nonsingular.
Solution: If A is not non-singular, then AX = O has
infinitely many solutions.
Thus, if AX = O fi X = O, then A must be non-singular.
Next, consider
(I + A)X = O
fi
AX = – X fi (AX) = (–X)¢
fi
X¢ A¢ = – X¢ fi X¢(–A) = –X¢
fi
X¢ A = X¢
fi X¢AX = X¢X
fi
X¢ (–X) = X¢X fi X¢X = O fi X = O.
\ I + A is non-singular.
LEVEL 2
Straight Objective Type Questions
Ê 1 2ˆ
Ê a 0ˆ
Example 86: Let A = Á
and B = Á
, a, b ΠN.
Ë 3 4˜¯
Ë 0 b˜¯
(a) Then there exists infinitely many B’s such that
AB = BA
(b) there cannot exist B such that AB = BA
(c) there exist more than one but finite number of B’s
such that AB = BA
(d) there exists exactly one B such that AB = BA
Ans. (a)
Solution: If b = a, B = aI
matrix A.
È5
Example 87: Let A = Í0
Í
| a | equals
ÎÍ0
and B commutes with every
(a) 52
(c) 1/5
Ans. (c)
(b) 1
(d) 5
5a
a
0
a˘
5a ˙ . If | A2 | = 25 then
˙
5 ˚˙
\
2
(25a) = 25 fi 25a = 1
fi
| a | = 1/5
Ê cos a - sin a ˆ
Example 88: If A = Á
and A + A¢ = I,
Ë sin a cos a ˜¯
then a equals
p
(a) 2np ± , n Œ I
2
(c) 2np ±
Ans. (b)
2p
,nŒI
3
fi
È1 0 ˘ È2 cos a
Í0 1 ˙ = Í 0
Î
˚ Î
fi
2 cos a = 1
fi
p
(b) 2np ± , n Œ I
3
(d) 2np ±
4p
,nŒI
3
0 ˘
2 cos a ˙˚
fi
cos a =
a = 2np ±
p
,nŒI
3
1
p
= cos
2
3
0
- tan(a / 2)˘
È
Example 89: Let A = Í
˙
0
Îtan(a / 2)
˚
a π (2n + 1) p, n Œ I, (I + A) (I – A)–1 equals
Ècos a
(a) Í
Î sin a
Ètan a
(c) Í
Î 0
Ans. (a)
Solution: | A2 | = 25 fi | A |2 = 25
2
Solution:
- sin a ˘
cos a ˙˚
È cos a
(b) Í
Î- sin a
0 ˘
tan a ˙˚
Ètan a
(d) Í
Î 0
sin a ˘
cos a ˙˚
0 ˘
- tan a ˙˚
1
tan(a / 2)˘
È
Solution: I – A = Í
˙
1
Î- tan(a / 2)
˚
then
(I – A)– 1 =
- tan(a / 2)˘
1
È
Ítan (a / 2)
˙
1
sec (a / 2) Î
˚
1
2
Now, (I + A) (I – A)–1
=
- tan(a / 2)˘
1
È
Ítan (a / 2)
˙
1
sec (a / 2) Î
˚
1
2
1
- tan(a / 2)˘
È
Ítan (a / 2)
˙
1
Î
˚
Matrices 5.23
A3 = A2 A
and
È1 - tan 2 (a / 2) -2 tan(a / 2) ˘
˙
Í
1 + tan 2 (a / 2) 1 + tan 2 (a / 2) ˙ Ècos a
Í
=
=
Í 2 tan (a / 2) 1 - tan 2 (a / 2) ˙ ÍÎ sin a
˙
Í
ÍÎ1 + tan 2 (a / 2) 1 + tan 2 (a / 2) ˙˚
0 ˘ È1 0 0 ˘
0˘
È1 0
È1 0
Í
˙
Í
˙
Í
= 0 -1 5 0 1 1 = 0 -11 19 ˙
Í
˙Í
˙
Í
˙
ÍÎ0 -10 14 ˙˚ ÍÎ0 -2 4 ˙˚
ÍÎ0 -38 46 ˙˚
- sin a ˘
cos a ˙˚
Now,
È 3 /2
Example 90: Let P = Í
ÍÎ -1/ 2
2015
P is
Q = PAP¢, then P¢Q
È1 -2015˘
1 ˙˚
1/ 2 ˘
˙, A =
3 / 2 ˙˚
È2015
(a) Í
Î0
(b) Í
Î 0
È1 2015˘
(c) Í
1 ˙˚
Î0
Ans. (c)
6I
6
6
6
d
È1 1˘
Í0 1˙ and
Î
˚
1 ˘
2015˙˚
È2015 2015˘
(d) Í
2015˙˚
Î 0
fi
Example 92: If a, b, c are non-zero, then number of
solutions of
2 x2
Ècos(p / 6) - sin (p / 6)˘
P¢ = Í
˙
Î sin (p / 6) cos(p / 6) ˚
Since
We have
È1 0 ˘
PP¢ = Í
fi P¢ = P –1
˙
Î0 1 ˚
Q = PAP¢ = PAP –1
Q2015 = (PAP –1)2015 = PA2015 P –1
fi
Thus,
is
Thus,
A
2015
È1 2015˘
= I + 2015B = Í
1 ˙˚
Î0
È1 0 0 ˘
Example 91: If A = Í0 1 1 ˙ , 6A–1 = A2 + cA + dI,
Í
˙
ÍÎ0 -2 4 ˙˚
then (c, d) is
(a) (– 6, 11)
(c) (11, 6)
Ans. (a)
(b) (–11, 6)
(d) (6, 11)
Solution: 6A–1 = A2 + cA + dI
fi
6I = A3 + cA2 + dA
We have
0˘
È1 0 0 ˘ È1 0 0 ˘ È1 0
2
Í
˙
Í
˙
Í
A = 0 1 1 0 1 1 = 0 -1 5 ˙
Í
˙Í
˙ Í
˙
ÎÍ0 -2 4 ˙˚ ÍÎ0 -2 4 ˙˚ ÍÎ0 -10 14 ˙˚
(1)
2
2
2
– x + 2y - z = 0
a 2 b2 c 2
(2)
2
2
2
– x - y + 2z = 0
a 2 b2 c 2
(3)
b2
-
(b) 8
(d) infinite
Solution: From (2), (3), we get
3 y2
= (P –1 P) A2015 (P –1P)
È0 1 ˘
A = I + B where B = Í
˙
Î0 0 ˚
Since, B 2 = O, we get B r = O " r ≥ 2.
y2
-
(a) 6
(c) 9
Ans. (d)
P¢Q2015P = P –1 (PA2015 P –1)P
Now,
z2 = 0
c2
a2
È cos(p / 6) sin (p / 6) ˘
Solution: P = Í
˙
Î- sin (p / 6) cos(p / 6)˚
fi
= A3 + cA2 + dA
= 1 + c + d, 0 = 19 + 5c + d
= –11 – c + d
= 46 + 14c + 4d, 0 = –38 – 10c – 2d
= 11, c = –6
b2
-
3z 2 = 0
c2
fi
y2 = z 2
b2
c2
Putting this in (1) we get
2 x2
a2
\
fi
where
-
2 y2 = 0
b2
fi
x2
a2
=
y2
b2
x 2 = y 2 z 2 = k2 (say)
=
a2
b2 c 2
x = ± ka, y = ± kb, z = ± kc,
k ΠR.
Example 93: If A and B are two matrices such that AB =
B and BA = A, then A2 + B2 is equal to
(a) 2AB
(b) 2BA
(c) A + B
(d) AB
Ans. (c)
Solution: We have
A2 + B2 = (BA)2 + (AB)2
= (BA) (BA) + (AB) (AB)
= B(AB)A + A(BA)B
= B(BA) + A(AB) = BA + AB = A + B.
5.24
Complete Mathematics—JEE Main
È2 0 1 ˘
Example 94: If A = Í2 1 3˙ , then A2 – 5A + 6I is
Í
˙
ÍÎ1 -1 0 ˙˚
equal to
È 1 -1 -3 ˘
(a) Í -1 -1 -10 ˙
Í
˙
ÍÎ-5 4
4 ˙˚
(c) O
Ans. (a)
1 -5˘
È1
Í
(b) -1 -1 4 ˙
Í
˙
ÍÎ-3 -10 4 ˙˚
(d) I
È2 0 1 ˘ È2 0 1 ˘ È5 -1 2 ˘
Solution: A = Í2 1 3˙ Í2 1 3 ˙ = Í9 -2 5 ˙
˙
Í
˙Í
˙ Í
ÍÎ1 -1 0 ˙˚ ÍÎ1 -1 0 ˙˚ ÍÎ0 -1 -2 ˙˚
2
Now, A2 – 5A + 6I
È2 0 1 ˘
È1 0 0 ˘
È5 -1 2 ˘
Í
˙
Í
˙
Í
˙
= Í9 -2 5 ˙ - 5 Í2 1 3 ˙ + 6 Í0 1 0 ˙
ÍÎ0 -1 -2 ˙˚
ÍÎ1 -1 0 ˙˚
ÍÎ0 0 1 ˙˚
È5 - 10 + 6 -1 - 0 + 0 2 - 5 + 0 ˘
= Í9 - 10 + 0 -2 - 5 + 6 5 - 15 + 0 ˙
Í
˙
ÍÎ 0 - 5 + 0 -1 + 5 + 0 -2 - 0 + 6 ˙˚
È 1 0 0˘
(c) Í- a 0 0 ˙
(d) none of these
Í
˙
ÍÎ ac b 1 ˙˚
Ans. (a)
Solution: Using the formula for inverse of a 3 × 3 triangular matrix given in theory, A–1 is the matrix given in (a).
È3 -3 4 ˘
Example 96: If A = Í2 -3 4 ˙ then A–1 is
Í
˙
ÍÎ0 -1 1 ˙˚
(a) A
(b) A2
3
(c) A
(d) A4
Ans. (c)
Solution: To show that A–1 = B ¤ AB = I. We have
È3
A = Í2
Í
ÍÎ0
È3
2
A(A ) = Í2
Í
ÍÎ0
2
4 ˘ È 3 -4 4 ˘
4 ˙ = Í 0 -1 0 ˙
˙
˙ Í
1 ˙˚ ÍÎ-2 2 -3˙˚
4 ˘ È 1 * *˘
0 ˙ = Í-2 * *˙
˙ Í
˙
-3˙˚ ÍÎ * * *˙˚
TIP
È 1 -1 -3 ˘
Í
˙
= Í -1 -1 -10 ˙
ÍÎ-5 4
4 ˙˚
[Need not evaluate the remaining terms as A3 π I3 ]
Next
È3 -4 4˘È3 -4 4˘
A = (A ) (A ) = Í 0 -1 0 ˙ Í 0 -1 0 ˙
˙
Í
˙Í
ÍÎ-2 2 -3˙˚ ÍÎ-2 2 -3˙˚
È1 0 0˘
Example 95: The inverse of the matrix A = Ía 1 0 ˙
Í
˙
is
ÍÎb c 1 ˙˚
0 0˘
È 1
Í
(a)
-a
1 0˙
Í
˙
ÍÎac - b -c 1 ˙˚
-3 4 ˘ È3 -3
-3 4 ˙ Í2 -3
˙Í
-1 1 ˙˚ ÍÎ0 -1
-3 4 ˘ È 3 -4
-3 4 ˙ Í 0 -1
˙Í
-1 1 ˙˚ ÍÎ-2 2
4
2
2
È1 0 0 ˘
= Í0 1 0 ˙ = I 3
Í
˙
ÍÎ0 0 1 ˙˚
0 0˘
È1
Í
(b) - a 0 0 ˙
Í
˙
ÍÎ b -c 1 ˙˚
A–1 = A3.
Thus,
EXERCISES
Concept-based
Straight Objective Type Questions
1. Let A be a 2 × 2 invertible matrix. For which of the
following functions det (f (A)) = f (det(A)) is not true?
(a) f (x) = x3
(b) f (x) = x–1, x π 0
(c) f (x) = 1 + x
(d) f (x) = x– 3, x π 0
Èa b ˘
2. Suppose a, b, c, d Œ C, and let A = Í
˙ . Then
Îc d ˚
which one of the following is not true?
(a) |det(A)| £ |a| + |b| + |c| + |d|
(b) |det(A)| £ (|a| + |b|) (|c| + |d|)
(c) |det(A)| £ (|a| + |c|) (|b| + |d|)
(d) |det(A)| £ | a |2 + | b |2 | c |2 + | d |2
3. Let A, B be two 3 × 3 matrices with entries from
real numbers. Which one of the following is true?
Matrices 5.25
(a)
(b)
(c)
(d)
(A + B)3 = A3 + 3 A2B + 3AB2 + B3
(AB)2 = O fi AB = O
(A + B) (A – B) = A2 – B2
(A + B)A = BA + A2
(a) C¢ = B¢ – D¢
(c) C¢ = B¢ + D¢
(b) C + D = B
(d) none of these
7. Suppose A, B are two 3×3 matrices such that A–1
exists. Then (A – B)A–1 (A + B) is equal to
4. Suppose A and B are two 3×3 matrices with entries
from complex numbers such that ABA = I. Which
one of the following is not true?
(a) B is invertible
(b) B –1 = A2
(c) A is not invertible (d) A4B2 = I
5. Let A be a 3×3 matrix with entries from the set of
real numbers. Suppose the equation AX = B has a
solution for every 3×1 matrix B with entries from
the set of real numbers. Then
(a) A¢ Y = B has no solution
(b) A¢ Y = O fi Y = O
(c) AX = O has a non-trivial solution
(d) A is a singular matrix
È cos q i sin q ˘
6. Suppose A = Í
˙ for some q Œ R. Let
Îi sin q cos q ˚
B, C, D be three 2 × 2 matrices such that AB = AC
– AD, then
(a) (A + B)A–1) (A – B) (b) A–1 B + B2
(c) (I – BAB –1) (A – B) (d) (I + BAB –1) (A + B)
8. Let A be 3×3 matrix such that A is orthogonal and
idempotent, then
(a) A must be symmetric
(b) det(A) = –1
(c) A + A–1 = 1
(d) none of these
9. Suppose A and B are two orthogonal matrices of
the same size such that det(A) + det(B) = 0. Then
(a) A + B = –I
(b) A + B = I
(c) det(A + B) = 0
(d) A + B = O
Ê a bˆ
, where a, b, c, d Œ R. ad π 0
10. Let A = Á
Ë c d ˜¯
If (a + d)A – A2 = A¢, then
(a) a = d
(c) a + d = 0
(b) a = d = 1
(d) a + d = 1
LEVEL 1
Straight Objective Type Questions
Ê 1 kˆ
11. Let Sk = Á
, k Œ N. Then (S2)n (Sk)–1
Ë 0 1˜¯
(where n ΠN) is equal to:
(a) S2n + k
(b) S2n – k
(c) S n
2 + k -1
(d) S n
2 -k
Èa b ˘
12. Let S be the set of all 2 × 2 real matrices A = Í
˙
Îc d ˚
2
such that a + d = 3 and A¢ = A – 3A. Then
(a) S contains infinite number of elements
(b) S = Q
(c) S contains exactly two elements
(d) S contains exactly 24 elements
13. Let A = (aij)3×2 be a 3×2 matrix with real entries
and B = AA¢. Then
(a)
(b)
(c)
(d)
B –1 is a 3 × 3 matrix
B –1 is a 2 × 2 matrix
B –1 does not exist
B –1 exists if and only if exactly one row of A
consists of zeros
14. Let A = (aij)3×3 be a matrix with aij Œ C. Let B be
a matrix obtained by interchanging two columns of
A. Then det(A + B) is equal to
(a) det(A) + det(B)
(b) 0
(c) 2 det(A)
(d) det(A) – det(B)
Ê 1 0ˆ
Ê 0 1ˆ
,J= Á
, and
15. If I = Á
˜
Ë 0 1¯
Ë -1 0˜¯
Ê cos q
B= Á
Ë - sin q
sin q ˆ
, then B equals
cos q ˜¯
(a) (cos q ) I + (sin q )J
(b) (sin q)I + (cos q )J
(b) (cos q )I – (sin q)J
(d) – (cos q) I + (sin q )J
16. If A is both diagonal and skew-symmetric, then
(a) A is a symmetric matrix
(b) A is a null matrix
(c) A is a unit matrix
(d) none of these matrix
5.26
Complete Mathematics—JEE Main
17. If A2 – 3A + 2I = 0, then A–1 equals
1
(A – 3I)
2
(c) A + 3I
(a)
1
(3I – A)
2
(d) none of these
(b)
18. If A is a square matrix of order 3 such that A2 =
2A, then |A|2 is equal to
(a) 2 |A|
(c) 16 |A|
(b) 8 |A|
(d) 0
19. If A is a square matrix then which one of the following is not a symmetric matrix
(a) A + A¢
(c) A¢A
(b) AA¢
(d) A – A¢
20. If A = (aij)3 × 3 where aij = cos (i + j), then
(a)
(b)
(c)
(d)
A is symmetric
A is skew-symmetric
A is a triangular matrix
A is a singular matrix
21. If A = (aij)3 × 3 is a matrix satisfying the equation
x3 – 3x + 1 = 0, then
(a)
(b)
(c)
(d)
A is a unit matrix
A is singular matrix
A is non-singular matrix
none of these
22. If A and B are two square matrices of the same size,
then (A + B)2 = A2+ 2AB + B2 can hold if and only
if
(a) AB = BA
(c) |AB| π 0
(b) AB + BA = O
(d) |AB| = 0
2 ˘
Èi 0 ˘
Èi
+X= Í
23. If Í
˙
˙ – X, then X is equal
Î3 -i ˚
Î3 4 + i ˚
to
1 ˘
È0 -1˘
È0
(b) Í
(a) Í
˙
˙
Î3 i ˚
Î0 2 + i ˚
0 ˘
È1
(c) Í
˙
Î0 2 - i ˚
2 ˘
Èi
(d) Í
˙
Î0 2 + i ˚
È0 -i ˘
È1 0 ˘
, B= Í
24. If A = Í
˙
˙ , then AB + BA is
Îi 0 ˚
Î0 -1˚
(a) null matrix
(c) invertible matrix
(b) unit matrix
(d) none of these
È1 2 3˘
25. A = Í 1 2 3 ˙ , then A is a nilpotent matrix of
Í
˙
ÍÎ-1 -2 -3˙˚
index
(a) 2
(b) 3
(c) 4
(d) 5
26. If A is a 2 ¥ 2 unitary matrix, then |A| is equal to
(a) 1
(c) + 1
27. If A =
(b) – 1
(d) none of these
1 Ê -1 - 3 ˆ
, then A–1 – A2 is equal to a
2 ÁË 3 -1 ˜¯
(a) null matrix
(c) unit matrix
(b) invertible matrix
(d) none of these
28. If C is a 3 × 3 matrix satisfying the relation
C2 + C = I, then C –2 is given by
(a) 2 C
(c) C
(b) 3 C – I
(d) 2I + C
29. If A, B and C are three square matrices of the same
size such that B = CA C –1, then CA3C –1 is equal
to
(a) B
(c) B3
(b) B2
(d) B9
30. If X is a 2 × 3 matrix such that |X ¢X| π 0, and A
= I – X(X ¢X)–1 X¢ then A2 is equal to
(a) A
(c) A–1
(b) I
(d) none of these
Ê p - qˆ
is orthogonal if and only
31. The matrix A = Á
Ë q p ˜¯
if
(a) p2 + q2 = 1
(c) p2 = q2 + 1
(b) p2 = q2
(d) none of these
32. The values of l for which the matrix
Êl 0 l ˆ
A = Á l 0 - l ˜ is orthogonal is
Á
˜
Ë0 1 0 ¯
(a) + 1
(b) ±1/ 3
(c) + 1/2
(d) ±1/ 2
33. The values of a for which the matrix
Ê a
a2 - 1
-3 ˆ
˜
Á
2
2
a + 4˜ is symmetric are
A = Áa +1
Á -3
-1 ˜¯
4a
Ë
(a) –1
(c) 3
3
Ê1
5
34. Let At = Á 2
Á
Ë4 7 - t
which inverse of At
(a) – 2, 1
(c) 2, – 3
(b) –2
(d) 2
2ˆ
t ˜ , then the value(s) of t for
˜
- 6¯
does not exist.
(b) 3, 2
(d) 3, – 1
Matrices 5.27
È a + ib c + id ˘
2
2
2
2
35. If A = Í
˙ , where a + b + c + d
c
+
id
a
ib
Î
˚
–1
= 1, then A is equal to
Èa - ib -c + id ˘
È a - ib
(b) Í
(a) Í
˙
Îc + id a + ib ˚
Î-c - id
c - id ˘
a + ib ˙˚
Èa - ib -c - id ˘
(c) Í
˙ (d) none of these
Îc - id a + ib ˚
1 ix
È 1 ix
- ix
e - e-ix
Í2 e + e
2
36. If A = Í
Í 1 eix - e-ix 1 eix + e-ix
ÍÎ 2
2
(a) for all real x
(b) for positive real x only
(c) for negative real x only
(d) none of these
(
) (
)
(
) (
)
È ab
37. If A = Í
ÍÎ- a 2
˘
˙
–1
˙ then A exists
˙
˙˚
b2 ˘
2
˙ , then A is equal
- ab ˙˚
(a) O
(c) – I
(b) I
(d) none of these
38. If A is 2 × 2 matrix such that A2 = O, then tr (A)
is
(a) 1
(c) 0
(b) – 1
(d) none of these
a b˘
39. If A = ÈÍ
˙ such that A satisfies the relation
Îc d ˚
A2 – (a + d)A = O, then inverse of A is
(a) I
(c) (a + d)A
(b) A
(d) none of these
3 2˘
40. If A = ÈÍ
, then A–3 is
˙
Î0 1 ˚
1 È1 -26 ˘
(a)
27 ÍÎ0 -27 ˙˚
(c)
1
27
È1 -26 ˘
Í0 27 ˙
˚
Î
43.
44.
45.
46.
(b) I
(d) A2
È1 2 2 ˘
If 3A = Í2 1 -2 ˙ and AA¢ = I, then x + y is
Í
˙
ÍÎ x 2 y ˙˚
equal to
(a) – 3
(b) – 2
(c) – 1
(d) 0
If the system of equations ax + y= 3, x + 2y = 3,
3x + 4y = 7 is consistent, then value of a is given
by
(a) 2
(b) 1
(c) – 1
(d) 0
If the system of equations x + 2y – 3z = 1, (p +
2)z = 3, (2p + 1) y + z =2 is inconsistent, then the
value of p is
(a) – 2
(b) – 1/2
(c) 0
(d) 2
The system of linear equations x + y + z = 2, 2x +
y – z = 3, 3x + 2y + kz = 4 has a unique solution
if
(a) k π 0
(b) – 1 < k < 1
(c) – 2 < k < 2
(d) k = 0
x + 2˘
È 4
47. If A = Í
˙ is an invertible matrix, then
Î2 x - 3 x + 1 ˚
x cannot take value
(a) – 1
(c) 3
(b) 2
(d) none of these
48. If A and B are two square matrices of the same
order, then which of the following is true.
(a) (AB)¢ = A¢ B¢
(b) (AB)¢ = B¢ A¢
(c) |AB| = 0 fi |A| = 0 and |B| = 0
1 È-1 -26 ˘
(b)
27 ÍÎ 0 -27 ˙˚
(d)
(a) O
(c) A
1
27
È-1 26 ˘
Í 0 -27˙
˚
Î
41. If A is a skew Hermitian matrix, then the main
diagonal elements of A are all
(a) purely real
(b) positive
(c) negative
(d) purely imaginary
È1 2 1 ˘
42. If A = Í0 1 -1˙ then A3 – 3A2 – A + 9I is equal
Í
˙
ÍÎ3 -1 1 ˙˚
to
(d) none of these
49. The values of a for which the system of equations
x + y + z = 1, x + 2y + 4z = a , x + 4y + 10z
= a 2 is consistent, are given by
(a) 1, – 2
(b) – 1, 2
(c) 1, 2
(d) none of these
50. The system of homogeneous equations
tx + (t + 1) y + (t – 1) z = 0
(t + 1) x + ty + (t + 2) z = 0
(t – 1) x + (t + 2) y + tz = 0
has a non-trivial solution for
(a)
(b)
(c)
(d)
three values of t
two values of t
one value of t
infinite number of values of t
5.28
Complete Mathematics—JEE Main
51. If A and B are two 3 ¥ 3 matrices and |A| π 0, then
which of the following are not true?
(a) 128 B
(c) 16 B
(b) 32 B
(d) 64 B
3 - i -i ˆ
Ê 2
Á
53. If A = 3 + i p
7 + i˜ , then A is
Á
˜
Ë i
7-i
e ¯
(a) |AB| = 0 fi |B| = 0
(b) |AB| π 0 fi |B| π 0
(c) |A–1| = |A|–1
(d) |A + A| = 2|A|
(a) symmetric
(c) skew Hermitian
Ê i -i ˆ
Ê 1 -1ˆ
and B = Á
, then A8 equals
52. If A = Á
˜
Ë -i i ¯
Ë -1 1 ˜¯
(b) Hermitian
(d) none of these
Assertion-Reason Type Questions
Statement-2: If | A| π 0, the system of equations
AX = B is consistent.
54. Statement-1: If A and B are two 3 ¥ 3 matrices
such that AB = O, then A = O or B = O.
Statement-2: If A, B and X are three 3 ¥ 3 matrices
such that AX = B, |A| π 0, then X = A–1 B.
1 100p ˘
È1 p ˘
, then A100 = ÈÍ
55. Statement-1: If A = Í
.
˙
1 ˙˚
Î0 1 ˚
Î0
Statement-2: If B is a 2 ¥ 2 matrix such that
B2 = O, then (I + B)n = I + nB for each n ΠN.
È2 1 1 ˘
56. Statement-1: As A = Í0 1 0 ˙ satisfies the equation
Í
˙
ÎÍ1 1 2 ˚˙
x3 – 5x2 + 7x – 3 = 0, then A is invertible.
Statement-2: If a square matrix A satisfies the
equation a0x n + a1 x n – 1 + º + an – 1x + an = 0,
and an π 0, then A is invertible.
57. Statement-1: If a, b, c are in A.P. the system of
equations
3x + 4y + 5z = a
(1)
4x + 5y + 6z = b
(2)
5x + 6y + 7z = c
(3)
is consistent.
Èa
58. Let A = Í 11
Îa21
a12 ˘
È x1 ˘
È y1 ˘
, X = Í ˙, Y = Í ˙
˙
a22 ˚
Î x2 ˚
Î y2 ˚
Statement-1: If A is symmetric, then X ¢AY = Y ¢AX
for each pair of X and Y.
Statement-2: If X¢AY = Y¢AX for each pair of X
and Y, then A is symmetric.
59. For any n ≥ 2. Let Mn (R) denote the set of all.
n ¥ n matrices over the set of real numbers.
Statement-1: If A Œ Mn (R), A π O with
det (A) = 0, then det (Adj A) = 0
Statement-2: For A ΠMn (R),
det (Adj A) = (det A)n–1
60. Let B be a 3 × 3 matrix such that (I – B)–1 exists.
Statement-1: If BX = X then X = O
Statement-2: I + B + B2 + …+ Bk
= (I – Bk+1) (I – B)–1 for all k Œ N.
LEVEL 2
Straight Objective Type Questions
Ê 1
61. If Á
Ë tan q
then
(a)
(b)
(c)
(d)
- tan q ˆ Ê 1
1 ˜¯ ÁË - tan q
a=b=1
a = cos2q, b = sin2q
a = sin2q, b = cos2q
a = 1, b = sin2q
tan q ˆ -1 Ê a -bˆ
=Á
,
Ë b a ˜¯
1 ˜¯
62. If a, b, c π 0 and a + b + c = 0, then the matrix
˘
È 1
1
1 ˙
Í1 + a
Í
˙
1
1+
1 ˙
Í 1
b
Í
1˙
Í 1
1
1+ ˙
ÍÎ
c ˙˚
is
Matrices 5.29
(a)
(b)
(c)
(d)
singular
non-singular
skew-symmetric
orthogonal
(a) A = O
(c) A is orthogonal
70 Suppose A and B are two 3 ¥ 3 non-singular matrices such that
63. Suppose matrix A satisfies the equation A2 – 5A +
7I = O. If A8 = aA + bI, then value of a is
(a) 1265
(c) –2599
(b) 2599
(d) 0
64. Let a, b, g be three real numbers and
1
cos(b - a ) cos(g - a )˘
È
Í
1
cos(g - b )˙
A = cos(a - b )
Í
˙
ÍÎ cos(a - g ) cos(b - g )
˙˚
1
then
(a) A is singular
(b) A is non-singular
(c) A is orthogonal
(d) none of these
Ê sin q
65. Let A(q) = Á
Ë i cos q
(a)
(b)
(c)
(d)
i cos q ˆ
, then
sin q ˜¯
A(q)–1 = A(–q)
A(q)–1 = A(p – q)
A(q)–1 does not exist
A(q)2 = A(2q)
66. Let A and B be two matrices such that AB = BA,
then which one of the following does not hold for
each n ΠN.
(a)
(b)
(c)
(d)
(b) 2A = I
(d) none of these
An B = BAn
(AB)n = AnBn
A nB = B nA
A nB n = B nA n
È1 2 1 ˘
67. Let A = Í0 1 -1˙ . The sum of all values of l
Í
˙
ÍÎ3 1 1 ˙˚
for which there exists a column vector X π O such
that AX = lX, is
(a) 0
(b) 1
(c) 2
(d) 3
68. Let a, b, c ΠR be such that a + b + c > 0 and
abc = 2. Let
Èa b c ˘
Í
A = b c a˙
Í
˙
ÎÍ c a b ˙˚
If A2 = I, then value of a3 + b3 + c3 is
(a) 7
(b) 2
(c) 0
(d) –1
69. If A is a 3 ¥ 3 skew-symmetric matrix with real
entries and trace of A2 equals zero, then
(AB)k = AkBk
for k = 2015, 2016, 2017, then
(a) AB = O
(b) BA = O
(c) AB = BA
(d) AB + BA = O
71. Let A be a square matrix of order 3 such that
|Adj A | = 100, then | A | equals
(a) ±10
(b) –100
(c) 100
(d) 25
72. Let M be a 3 ¥ 3 matrix satisfying
È0 ˘ È – 1˘
È 1 ˘ È 1˘
È1˘ È 0 ˘
Í
˙
Í
˙
Í
˙
Í
˙
M 1 = 2 , M – 1 = 1 , and M Í1˙ = Í 0 ˙
Í ˙ Í ˙
Í ˙ Í ˙
Í˙ Í ˙
ÍÎ0 ˙˚ ÍÎ 3 ˙˚
ÍÎ 0 ˙˚ ÍÎ – 1˙˚
ÍÎ1˙˚ ÍÎ12 ˙˚
then sum of the diagonal entries of M is
(a) 0
(c) 6
(b) –3
(d) 9
73. If A, B and A + B are non-singular matrices, then
(A–1 + B –1) [A – A (A + B)–1 A]
equals
(a) O
(c) A
(b) I
(d) B
74. If A + B is a non-singular matrix, then
A – B – A (A + B)–1A + B (A + B)–1 B
equals
(a) O
(c) A
(b) I
(d) B
75. If A and B are two matrices such that
A + B = AB, then
(a) A = I
(b) B = I
(c) AB = BA
(d) AB = I
a
76. Let A = ÊÁ
Ëc
(a) ad
(c) 1
Ê0
77. Let A = Á y
Á
Ë0
bˆ
such that A3 = O, then a + d equals
˜
d¯
(b) bc
(d) 0
x 0ˆ
0 - x ˜ , then A3 equals
˜
y 0¯
(a) O
(c) (x2 + y2) I
(b) x2I
(d) none of these
5.30
Complete Mathematics—JEE Main
Previous Years' AIEEE/JEE Main Questions
Èa b ˘
Èa b ˘
1. If A = Í
and A2 = Í
˙ , then
˙
Îb a ˚
Îb a ˚
(a)
(b)
(c)
(d)
a = a2 + b2, b = 2ab
a = a 2 + b 2, b = a 2 – b 2
a = 2ab, b = a2 + b2
a = a2 + b2, b = 2ab
Ê0
2. Let A = Á 0
Á
Ë -1
about matrix
Ê 1 2ˆ
Ê a 0ˆ
and B = Á
a, b ΠN. Then
7. Let A = Á
˜
Ë 3 4¯
Ë 0 b˜¯
[2003]
0 -1ˆ
-1 0 ˜ . The only correct statement
˜
0 0¯
A is
A–1 does not exist
A = (–1)I, where I is a unit matrix
A is a zero matrix
[2004]
A2 = I
1
1
1
4
2
2ˆ
Ê
Ê
ˆ
3. Let A = Á 2 1 -3˜ and (10)B = Á -5 0 a ˜ .
Á
Á
˜
˜
Ë1 1 1 ¯
Ë 1 -2 3 ¯
(a)
(b)
(c)
(d)
If B is the inverse of matrix A, then a is
(a) 2
(c) –2
(b) –1
(d) 5
4. If A – A + I = O, then the inverse of A is
(b) I – A
(d) A
An = nA + (n – 1)I
An = (2n – 1) A + (n – 1)I
An = nA – (n – 1)I
An = (2n – 1) A – (n – 1)I
[2005]
[2005]
6. If A and B are square matrices of size n ¥ n such
that
A2 – B2 = (A – B) (A + B), then which of the following will be always true?
(a)
(b)
(c)
(d)
either A or B is an identity matrix
A=B
AB = BA
either A or B is a zero matrix
(a) 52
(c) 1/5
a˘
5a ˙ . If | A2 | = 25 then | a |
˙
5 ˙˚
(b) 1
(d) 5
[2007]
9. Let A be a 2 ¥ 2 matrix with real entries. Let I be
the 2 ¥ 2 identity matrix. Denote by tr(A), the sum
of diagonal entries of A. Assume that A 2 = I.
Statement-1: If A π I and A π – I, then
det (A) = – 1.
Statement-2: If A π I and A π – I, then tr(A) π O.
[2008]
10. Let A be a 2 ¥ 2 matrix.
È1 0 ˘
È1 0 ˘
and I = Í
5. If A = Í
˙
˙ , then which one of
Î1 1 ˚
Î0 1 ˚
the following holds for all n ≥ 1, by the principle
of mathematical induction
(a)
(b)
(c)
(d)
È5 5a
8. Let A = Í0 a
Í
ÍÎ0 0
equals
[2004]
2
(a) A – I
(c) A + I
(a) there exist infinitely many B’s such that AB = BA
(b) there cannot exist B such that AB = BA
(c) there exist more than one but finite number of
B’s such that AB = BA
(d) there exists exactly one B such that AB = BA
[2006]
Statement-1:
Statement-2:
adj (adj A) = A
|adj A| = |A|
11. Let A be a 2 ¥ 2 matrix with non-zero entries and
let A2 = I, where I is 2 ¥ 2 identity matrix. Define
Tr(A) = Sum of diagonal elements of A,
|A| = determinant of matrix A.
Statement-1: Tr (A) = 0
Statement-2: |A| = 1
[2010]
12. The number of 3 ¥ 3 non-singular matrices with
four entries as 1 and all other entries 0, is
(a) 6
(c) less than 4
(b) at least 7
(d) 5
13. Consider the system of linear equations
x1 + 2x2 + x3 = 3
2x1 + 3x2 + x3 = 3
3x1 + 5x2 + 2x3 = 1
[2006]
[2009]
The system has
[2010]
Matrices 5.31
(a)
(b)
(c)
(d)
a unique solution
no solution
infinite number of solutions
exactly 3 solutions
[2010]
14. The number of values of k for which the linear
equations
4x + ky + 2z = 0
kx + 4y + z = 0
2x + 2y + z = 0
possess a non zero solution is
(a) 0
(c) 2
(b) 3
(d) 1
[2011]
Èw 0 ˘
15. If w π 1 is cube root of unity and H = Í
˙
Î0 w ˚
then H70 equals to
(a) O
(c) H2
(b) –H
(d) H
[2011]
16. If trivial solution is the only solution of the system
of linear equations:
x – ky + z = 0
kx + 3y – kz = 0
3x + y – z = 0
then set of all values of k is
(a) R – {2, –3}
(c) R –{–3}
(b) R –{2}
(d) {2, –3}
[2011]
17. Let A and B be two symmetric matrices of order 3.
Statement-1: A(BA) and (AB)A are symmetric
matrices
Statement-2: AB is symmetric if matrix multiplication of A with B is commutative
[2011]
18. Let P and Q be 3 ¥ 3 matrices such that P π Q.
If P3 = Q3 and P2Q = Q2P, then det (P2 + Q2) is
equal to
(a) 1
(b) 0
(c) –1
(d) –2
[2012]
Ê 1 0 0ˆ
19. Let A = Á 2 1 0˜ . If u1 and u2 are column maÁ
˜
Ë 3 2 1¯
Ê 1ˆ
Ê 0ˆ
trices such that Au1 = Á 0˜ and Au2 = Á 1˜ , then
Á ˜
Á ˜
Ë 0¯
Ë 0¯
u + u is equal to
1
2
Ê -1ˆ
(a) Á 1˜
Á ˜
Ë -1¯
Ê 1ˆ
(c) Á -1˜
Á ˜
Ë -1¯
Ê -1ˆ
(b) Á -1˜
Á ˜
Ë 0¯
Ê -1ˆ
(d) Á 1˜
Á ˜
Ë 0¯
[2012]
È1 a
20. If P = Í1 3
Í
ÍÎ2 4
A and |A| = 4,
(a) 11
(c) 0
3˘
3 ˙ is the adjoint of a 3 × 3 matrix
˙
4 ˙˚
then a is equal to
(b) 5
(d) 4
[2013]
21. The number of values of k for which the system of
equations
(k + 1)x + 8y = 4 k
kx + (k + 3)y = 3 k – 1
has no solution, is
(a) 1
(b) 2
(c) 3
(d) infinite
[2013]
22. If the system of linear equations
x1 + 2x2 + 3x3 = 6
x1 + 3x2 + 5x3 = 9
2x1 + 5x2 + ax3 = b
is consistent and has infinite number of solutions,
then
(a) a = 8, b can be any real number
(b) b = 15, a can be any real number
(c) a ΠR Р{8} and b ΠR Р{15}
(d) a = 8, b = 15
[2013, online]
23. If p, q, r are 3 real numbers satisfying the matrix
equation,
È3 4 1 ˘
[ p q r ] Í3 2 3˙ = [3 0 1]
Í
˙
ÎÍ2 0 2 ˚˙
then 2p + q – r equal
(a) –3
(b) –1
(c) 4
(d) 2
[2013, online]
24. Consider the system of equations:
x + ay = 0, y + az = 0, and z + ax = 0.
Then the set of all values of a for which the system
has a unique solution is
(a) R – {1}
(b) R – {–1}
(c) {1, –1}
(d) {1, 0, –1}
[2013, online]
a12 ˆ
ÏÊ a
¸
25. Let S = ÌÁ 11
˜¯ aij Œ{0, 1, 2}, a11 = a22 ˝
Ë
a
a
Ó 21
˛
22
Then the number of non-singular matrices in the
set S is
(a) 27
(b) 24
(c) 10
(d) 20
[2013, online]
26. The matrix A2 + 4A – 5I, where I is the identity
È1 2 ˘
matrix and A = Í
˙ , equals:
Î4 -3˚
5.32
Complete Mathematics—JEE Main
È0 -1˘
(b) 4 Í
˙
Î2 2 ˚
È2 1 ˘
(a) 4 Í
˙
Î2 0 ˚
È2 1 ˘
(c) 32 Í
˙
Î2 0 ˚
È1 1 ˘
(d) 32 Í
˙
Î1 0 ˚
[2013, online]
27. Let A, other than I or –I, be a 2 × 2 real matrix
such that A2 = I, I being the unit matrix. Let Tr(A)
be the sum of diagonal elements of A.
Statement-1: Tr(A) = 0
Statement-2: det(A) = –1
[2013, online]
28. If A is an 3 × 3 non-singular matrix such that AA¢
= A¢A and B = A–1 A¢, then BB¢ equals:
(a) I
(b) B –1
(c) (B –1)¢
(d) I + B
[2014]
29. If B is a 3 × 3 matrix such that B2 = O, then
det [(I + B)50 – 50B] is equal to:
(a) 1
(c) 3
30. Let A be
È1
A Í0
Í
ÍÎ0
(b) 2
(d) 50
a3
2
2
1
× 3 matrix
3˘ È 0 0
3˙ = Í 1 0
˙ Í
1˙˚ ÍÎ0 1
[2014, online]
such that
1˘
0˙
˙
0 ˙˚
then A–1 is :
È3 1 2 ˘
(a) Í3 0 2 ˙
˙
Í
ÎÍ1 0 1 ˙˚
È 0 1 3˘
(c) Í0 2 3˙
Í
˙
ÍÎ1 1 1˙˚
È3 2 1 ˘
(b) Í3 2 0 ˙
˙
Í
ÎÍ1 1 0 ˙˚
È 1 2 3˘
(d) Í0 1 1˙
Í
˙
ÍÎ0 2 3˙˚
[2014, online]
È y˘
È1 2 x ˘
Í ˙
31. If A = Í
˙ and B = Í x ˙ be such that AB
Î3 -1 2 ˚
ÎÍ1 ˚˙
È6 ˘
= Í ˙ , then:
Î8 ˚
(a) y = 2x
(b) y = –2x
(c) y = x
(d) y = –x
[2014, online]
32. Let A and B be any two 3×3 matrices. If A is symmetric and B is skew-symmetric, then the matrix
AB – BA is:
(a) skew-symmetric
(b) symmetric
(c) neither symmetric nor skew-symmetric
(d) I or –I, where I is an identity matrix.
[2014, online]
2˘
È1 2
Í
33. If A = 2 1 -2 ˙ is a matrix satisfying the equa˙
Í
b ˙˚
ÍÎ a 2
tion AAT = 9I, where I is 3 ¥ 3 identity matrix, then
the ordered pair (a, b) is equal to:
(a) (2, –1)
(b) (–2, 1)
(c) (2, 1)
(d) (–2, –1)
[2015]
34. If A is 3 ¥ 3 matrix such that |5adj A| = 5, then |A|
is equal to:
1
(a) ±
(b) ± 5
5
1
[2015, online]
(c) ± 1
(d) ±
25
0 -1˘
35. If A = È
, then which one of the following
Í 1 0˙
Î
˚
statement is not correct?
(a) A4 – I = A2 + I
(b) A3 – I = A(A – I)
2
2
(c) A + I = A(A – I) (d) A3 + I = A(A3 – I)
[2015, online]
È5a -b ˘
36. If A = Í
˙ and Aadj A = AA¢, then 5a + b
Î3 2˚
is equal to
(a) –1
(b) 5
(c) 4
(d) 13
[2016]
È 3 1 ˘
Í
˙
È1 1˘
2
2 ˙
,A= Í
37. If P = Í
˙ and Q = PAP ¢, then
Í 1
3˙
Î0 1˚
Í˙
Î 2 2 ˚
P ¢Q2015 P is:
0 ˘
È2015
È0 2015˘
(a) Í
(b) Í
˙
0 ˚
2015˙˚
Î 1
Î0
È1 2015˘
(c) Í
1 ˙˚
Î0
1 ˘
È2015
(d) Í
2015˙˚
Î 0
[2016, online]
È-4 -1˘
38. If A = Í
˙ , then the determinant of the matrix
Î3 1˚
(A2016 – 2A2015 – A2014) is:
(a) –175
(b) 2014
(c) 2016
(d) –25
[2016, online]
39. Let A be a 3 ¥ 3 matrix such that A2 – 5A + 7I = 0.
1
(5I – A)
Statement-1: A–1 =
7
Statement-2: The polynomial A3 –2A2 – 3A + I can
be reduced to 5(A + 4I). Then
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement-1 is true, but Statement-2 is false.
(d) Statement-1 is false, but Statement-2 is true.
[2016, online]
Matrices 5.33
Previous Years' B-Architecture Entrance
Examination Questions
1. If A and B are square matrices of the same order
then which one of the following is always true?
(a) (A + B)–1 = A–1 + B –1
(b) adj (AB) = (adj B) (adj A)
(c) A and B are non-zero and |AB| = 0 ¤ |A| = 0 and
|B| = 0
[2006]
(d) (AB)–1 = A–1 B –1
È1 1 0 ˘
2. Let A = Í0 1 0 ˙ , and let I denote the 3 × 3
˙
Í
ÎÍ0 0 1 ˚˙
identity matrix. Then 2A2 – A3 =
(a) A + I
(b) A – I
(c) I – A
(d) A
[2008]
3. Let A and B be 2 × 2 matrices with real entries.
Let I be the 2 × 2 identity matrix. Denote by tr(A)
the sum of diagonal entries of A.
Statement-1: AB – BA π I
Statement-2: tr(A + B) = tr(A) + tr(B) and
tr(AB) = tr(BA)
[2009]
Èa b ˘
4. Let A = Í
˙ be a 2 × 2 real matrix. If A – a I
Îc d ˚
is invertible for every real number a, then
(a) bc > 0
(b) bc = 0
(d) a = 0
[2010]
(c) bc < min Ê0, 1 ad ˆ
Ë 2 ¯
5. Let A and B be two 2 × 2 matrices.
Statement-1:
Statement-2:
A (adj A) = |A| I2
adj(AB) = (adj A)(adj B)
Èa b ˘
6. Let A = Í
˙ , be a 2 × 2 matrix, where a, b,
Îc d ˚
c, d take values 0 or 1 only. The number of such
matrices which have inverses is:
(a) 8
(c) 7
(c) 6
(d) 5
[2012]
Èa b ˘
7. Let S be the set of all real matrices A = Í
˙
Îc d ˚
such that a + d = 2 and A¢ = A2 – 2A. Then S:
is an empty set
has exactly one element
has exactly two elements
has exactly four elements
(a) S2n + k –1
(b) S2n + k -1
(c) S2n - k
(d) S2n – k
1
9. Let A = ÈÍ
Î0
10
Adj(A ) = B,
[2014]
È b1 b2 ˘
1˘
10
and B = Í
˙ . If 10A +
˙
1˚
Îb3 b4 ˚
then b1 + b2 + b3 + b4 is equal to:
(a) 91
(c) 111
(b) 92
(d) 112
[2015]
È 1 -2 4˘
10. If for a matrix A, |A| = 6 and adj A = Í 4 1 1 ˙ ,
Í
˙
ÍÎ-1 k 0 ˙˚
then k is equal to
(a) 0
(b) 1
(c) 2
(d) –1
[2016]
Answers
Concept-based
1. (c)
5. (b)
9. (c)
2. (a)
6. (c)
10. (b)
3. (d)
7. (a)
4. (c)
8. (a)
Level 1
[2011]
(a)
(b)
(c)
(d)
È1 k ˘
8. Let Sk = Í
˙ k Œ N, where N is the set of
Î0 1 ˚
natural numbers, then (S2)n (Sk)–1, for n Œ N is:
[2013]
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
(b)
(a)
(d)
(b)
(a)
(a)
(c)
(d)
(a)
(d)
(d)
(a)
(a)
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
(b)
(b)
(a)
(a)
(d)
(d)
(a)
(c)
(a)
(b)
(a)
(a)
(b)
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
57.
(c)
(b)
(c)
(a)
(c)
(d)
(a)
(d)
(a)
(c)
(b)
(b)
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
(b)
(b)
(a)
(d)
(a)
(c)
(c)
(a)
(a)
(c)
(d)
(b)
5.34
Complete Mathematics—JEE Main
Level 2
5. Let X1, X2 X3 be solutions of AX = B, when
61. (b)
62. (b)
63. (a)
64. (a)
65. (b)
66. (c)
67. (d)
68. (a)
69. (a)
70. (c)
71. (a)
72. (d)
73. (b)
74. (a)
75. (c)
76. (d)
77. (a)
Previous Years' AIEEE/JEE Main Questions
1. (a)
2. (d)
3. (d)
4. (b)
5. (c)
6. (c)
7. (a)
8. (c)
9. (c)
10. (b)
11. (c)
12. (b)
13. (b)
14. (c)
15. (d)
16. (a)
17. (b)
18. (b)
19. (c)
20. (a)
21. (a)
22. (d)
23. (a)
24. (b)
25. (d)
26. (a)
27. (a)
28. (a)
29. (a)
30. (a)
31. (a)
32. (b)
33. (d)
34. (a)
35. (c)
36. (b)
37. (c)
38. (d)
39. (a)
Ê1 ˆ
B = Á 0˜ ,
Á ˜
Ë 0¯
2. (d)
6. (c)
10. (c)
3. (a)
7. (a)
4. (c)
8. (d)
Hints and Solutions
fi A¢ is invertible.
Thus A¢Y = O fi Y = O.
6. det(A) = cos2q + sin2q = 1 π 0
\ A–1 exist.
Now,
AB = A(C – D)
–1
fi
A (AB) = A–1 (A(C – D))
B=C–D fi C=B+D
fi
C¢ = (B + D)¢ = B¢ + D¢
7. (A – B) A–1 (A + B)
= (I – BA–1) (A + B)
= A + B – BA–1 A – BA–1 B
= A – BA–1 B
= A + B – BA–1 A – BA–1 B
= (A + B) – BA–1 (A + B)
= (A + B) (I – BA–1)
= (A + B)A–1 (A – B)
1. If f (x) = 1 + x, then det f (A) = 1 + det(A) may not
hold. Take A = –I.
2. |det(A)| = |ad – bc| £ |a||d| + |b||c|
£ (|a| + |b|) (|c| + |d|)
Also, |det(A)| £ (|a| + |c|) (|b| + |d|)
For truth of (c), see Example 6
[
È 5 -1˘ ˘
(a) does not hold in general. Take A = Í
˙˙
Î10 2 ˚ ˚
3. (a), (b), (c) cannot hold in general unless AB = BA.
4. det(ABA) = 1
fi
det(A) det(B) det(A) = 1
fi
det(A) π 0,
Both A
–1
det(B) π 0
and B
–1
and
\
AA¢ = A2
fi
A¢ = A
A2 = I
fi
A–1 (AA¢) = A–1A2
9. As A, B are orthogonal matrices, AA¢ = BB¢ = I,
det(A) π 0, det(B) π 0. Let C = A + B.
B2 = (A–2)2 = A– 4
exist
fi
fi AC ¢B = B + AB ¢B = B + A
Now,
det(A + B) = det(AC ¢B) = det(A) det(C ¢) det(B)
= det(A) det(C) (– det(A))
fi
det(C) = – (det(A))2 det(C) = – det(C)
fi
det(C) = 0
2
Ê a b ˆ Ê a b ˆ Ê a + bc b(a + d )ˆ
=
10. A2 = Á
Ë c d ˜¯ ÁË c d ˜¯ ÁË c(a + d ) bc + d 2 ˜¯
\
(a + d)A – A2 = A¢
fi
0 ˆ Ê a bˆ
Ê ad - bc
=
fi ad = a, d
ÁË 0
ad - bc˜¯ ÁË c d ˜¯
As ad π 0, we get a = d = 1.
Also ABA = I fi BA = A–1 fi B = A–1A–1 = A–2
fi
8. AA¢ = I
C ¢ = A¢ + B ¢ fi AC ¢ = AA¢ + AB ¢ = I + AB ¢
Concept-based
fi
Ê 0ˆ
Á 0˜ respectively. Let C be the 3 × 3
Á ˜
Ë1 ¯
matrix [X1 X2 X3], then AC = I fi A is invertible.
Previous Years' B-Architecture Entrance
Examination Questions
1. (b)
5. (c)
9. (d)
Ê 0ˆ
Á1 ˜ ,
Á ˜
Ë 0¯
A 4B 2 = I
Level 1
Ê 1 2ˆ
=I+B
11. S2 = Á
Ë 0 1˜¯
where
Matrices 5.35
\
Ê 1 0ˆ
Ê 0 2ˆ
and B = Á
I= Á
˜
Ë 0 1¯
Ë 0 0˜¯
Note that B2 = O, therefore, Br = O
14. In A + B two rows are identical.
" r ≥ 2.
(S2)n = (I + B)n = I + nB
Ê 1 2 nˆ
=Á
Ë 0 1 ˜¯
Thus,
Ê 1 -k ˆ
Sk-1 = Á
Ë 0 1 ˜¯
Also,
Ê 1 2 nˆ Ê 1 - k ˆ
\ (S2) (Sk) = Á
Ë 0 1 ˜¯ ÁË 0 1 ˜¯
n
Èa b ˘ Èa b ˘
12. A2 = Í
˙Í
˙
Îc d ˚ Îc d ˚
È a 2 + bc b(a + d )˘
= Í
˙
ÍÎc(a + d ) bc + d 2 ˙˚
Èa + bc
3b ˘
= Í
˙
ÍÎ 3c
bc + d 2 ˙˚
Èa 2 - 3a + bc
˘
0
˙
A – 3A = Í
2
ÍÎ
0
d - 3d + bc ˙˚
2
A¢ = A2 – 3A, we get
b = 0, c = 0,
fi a (a – 4) = 0
fi
a = a2 – 3a, d = d2 – 3d
and d(d – 4) = 0
a = 0 or a = 4 and d = 0 or d = 4
But then a + d cannot be 3.
\
I= A
{
}
1
(3I - A)
2
1
(3I – A)
2
18. | A2 | = | 2A | = 23 | A | fi | A |2 = 8 | A |
19. (A – A¢)¢ = A¢ – (A¢)¢ = A¢ – A = – (A – A¢)
\ A – A¢ is skew-symmetric matrix.
Ê cos 2 cos 3 cos 4ˆ
20. A = Á cos 3 cos 4 cos 5˜ = A¢
Á
˜
Ë cos 4 cos 5 cos 6¯
fi A –1 =
22. (A + B)2 = (A + B) (A + B) = A2 + BA + AB + B2
2
˘
0
Èa c ˘ Èa - 3a + bc
=
˙
Íb d ˙ Í
Î
˚ ÍÎ
0
d 2 - 3d + bc ˙˚
fi
fi
21. A3 – 3A + I = 0 fi I = A(3I – A2)
fi A is invertible and hence non-singular.
2
Since
15. B = (cos q)I + (sin q)J.
16. A is a diagonal matrix, aij = 0 " i π j
As A is a skew-symmetric matrix,
aii = 0 " i
\
A=O
17. A2 – 3A + 2I fi 2I = A(3I – A)
–1
Ê 1 2n - k ˆ
= S2 n - k
=Á
Ë0
1 ˜¯
\
B –1 does not exist.
S =f
13. Let
Ê a pˆ
A = Á b q ˜ , then
Á
˜
Ëc r¯
Ê a 2 + p2 ab + pq ac + pr ˆ
Á
˜
B = AA¢ = Á ab + pq b2 + q 2 bc + qr ˜
ÁË ac + pr bc + qr c 2 + r 2 ˜¯
a p 0 a b c
det (B) = b q 0 p q r = 0
c r 0 0 0 0
\ (A + B)2 = A2 + 2AB + B2 if and only if
AB = BA
2 ˘ È i 0 ˘ È0
2 ˘
Èi
-Í
=Í
23. 2X = Í
˙
˙
˙
Î3 4 + i ˚ Î3 -i ˚ Î0 4 + 2i ˚
fi
1 ˘
È0
X= Í
˙
Î0 2 + i ˚
È0 -i ˘ È1 0 ˘ È0 i ˘
24. AB = Í
˙Í
˙=Í
˙
Î i 0 ˚ Î0 -1˚ Î i 0 ˚
È1 0 ˘ È0 -i ˘ È 0 -i ˘
and BA = Í
˙=Í
˙Í
˙
Î0 -1˚ Î i 0 ˚ Î-i 0 ˚
Thus, AB + BA = O
È0 0 0 ˘
È 1 2 3 ˘ È 1 2 3˘
Í
˙
Í
˙
1 2 3 = Í0 0 0 ˙
25. A = 1 2 3
Í
˙Í
˙
Í
˙
ÍÎ-1 -2 -3˙˚ ÍÎ-1 -2 3˙˚
ÍÎ0 0 0 ˙˚
\ A is a nilpotent matrix of index 2
2
Èa
26. Let A = Í
Îc
AA* = I fi
b˘
Èa c ˘
, A* = Í
˙
d ˙˚
Îb d ˚
| A | | A*| = 1
fi (ad – bc) (a d - b c) = 1
fi | ad – bc |2 = 1
fi ad Рbc = eiq for some q ΠR.
5.36
Complete Mathematics—JEE Main
Ê -1 / 2 - 3 / 2ˆ Ê cos(4p / 3) sin (4p / 3) ˆ
27. A = Á
˜
˜ =Á
Ë 3 / 2 -1 / 2 ¯ Ë - sin (4p / 3) cos(4p / 3)¯
Note that A3 = I fi A2 – A–1 = O
28. C2 + C = I fi C(C + I) = I
fi C –1 = C + I
Thus, C –2 = (C + I)2 = C2 + 2C + I
= I – C + 2C + I = 2I + C
29. CA3C–1 = (CAC –1) (CAC –1) (CAC –1) = B3
30. A2 = (I – X(X¢X) –1 X¢)2
= I – 2X (X¢X)–1 X¢ + X(X¢X)–1 X¢ X(X¢X)–1 X¢
= I – 2X(X¢X)–1 X¢ + X(X¢X)–1 X¢
= I – X(X¢X)–1 X¢ = A
31. AA¢ = I
p - q ˆ Ê p q ˆ Ê 1 0ˆ
fi ÊÁ
=
Ë q p ˜¯ ÁË -q p˜¯ ÁË 0 1˜¯
Ê p2 + q 2
fi Á
Ë 0
ˆ Ê 1 0ˆ
˜
˜ =Á
p 2 + q 2 ¯ Ë 0 1¯
0
\ p2 + q2 = 1
32. AA¢ = I
Êl 0 l ˆ Êl
fi Á l 0 -l ˜ Á 0
Á
˜Á
Ë 0 1 0 ¯ Ël
Ê 2l 2
Á
fi Á 0
Á 0
Ë
\ 2l2 = 1
33. A¢ = A
0
2l
0
2
fi
l 0ˆ Ê 1 0 0ˆ
0 1˜ = Á 0 1 0˜
˜ Á
˜
- l 0¯ Ë 0 0 1¯
0 ˆ Ê 1 0 0ˆ
˜
0 ˜ = Á 0 1 0˜
Á
˜
1˜¯ Ë 0 0 1¯
fi a + 1 = a2 – 1 and 4a = a2 + 4
fi a + 1 = 4a – 4 – 1 fi 6 – 3a fi a = 2.
34. Since At–1 does not exist, | At | = 0
fi (–30 – t(7 – t)) – 3(–12 – 4t) + 2(14 – 2t – 20)
=0
fi t2 + t – 6 = 0 fi t = –3, 2
Èa - ib -c - id ˘
35. A–1 = Í
˙
Îc - id a + ib ˚
1
[(eix + e–ix)2 – (eix – e–ix)2]
4
1
(4eix e–ix) = 1 π 0
=
4
Thus, A–1 exists for all x Œ R.
b2 ˘ È ab
˙Í
- ab ˙˚ ÍÎ- a 2
Èa 2 b2 - a 2 b2
= Í
ÍÎ - a3 b + a3 b
b2 ˘
˙
- ab ˙˚
ab3 - ab3 ˘
˙ =O
- a 2 b2 + a 2 b2 ˙˚
Èa 2 + bc ab + bd ˘
Èa b ˘
2
38. Let A = Í
˙ , then A = ÍÍ ac + cd bc + d 2 ˙˙
Îc d ˚
Î
˚
Now, A2 = O
fi a2 + bc = 0, (a + d) b = 0
(a + d) c = 0, bc + d2 = 0
Suppose a + d π 0, then b = 0, c = 0.
\ a2 = 0, d2 = 0 fi a = 0, d = 0
Contradiction.
Thus, a + d = 0 we get tr(A) = 0
39. Suppose A–1 exists, then
A–1 A2 – (a + d)A–1 A = O
fi A – (a + d)I = O
È- d b ˘
fi Í
˙ =O
Î c -a˚
fi a = b = c = d = 0 fi |A| = 0
A contradiction. Thus, A–1 does not exist.
40. We have A–1 =
1 È1 -2 ˘
3 ÍÎ0 3 ˙˚
È1 -2 ˘ È1 -2 ˘ È1 -2 ˘
Í0 3 ˙ Í0 3 ˙ Í0 3 ˙
Î
˚Î
˚Î
˚
1
8
1
2
1 È
1 È1 -26 ˘
˘È
˘
=
=
Í
˙
Í
˙
27 Î0 9 ˚ Î0 3 ˚
27 ÍÎ0 27 ˙˚
fi A–3 =
l = ± 1/ 2
a2 - 1
a + 1 -3ˆ Ê a
-3 ˆ
Ê a
Á
˜
Á 2
2
4 a˜ = Á a + 1
2
a 2 + 4˜
fi Áa -1
˜
ÁË -3
-1 ˜¯
4a
a 2 + 4 -1˜¯ ÁË -3
36. | A | =
È ab
37. A2 = Í
ÍÎ- a 2
1
27
41. A* = –A = fi ( A)¢ = –A fi aii = –aii " i
fi 2Re(aii) = 0 fi aii is purely imaginary.
È1 2 1 ˘ È1 2 1 ˘ È 4
42. A = Í0 1 -1˙ Í0 1 -1˙ = Í-3
Í
˙Í
˙ Í
ÍÎ3 -1 1 ˙˚ ÍÎ3 -1 1 ˙˚ ÍÎ 6
È 4 3 0 ˘ È1 2 1 ˘ È 4
3
A = Í-3 2 -2 ˙ Í0 1 -1˙ = Í-9
Í
˙Í
˙ Í
ÍÎ 6 4 5 ˙˚ ÍÎ3 -1 1 ˙˚ ÍÎ 21
Now, A3 – 3A2 – A + 9I
È 4 11 1 ˘
È4 3 0˘
Í
˙
= -9 -2 -7 - 3 Í-3 2 -2 ˙
Í
˙
Í
˙
ÍÎ 21 11 7 ˙˚
ÍÎ 6 4 5 ˙˚
2
3 0˘
2 -2 ˙
˙
4 5 ˙˚
11 1 ˘
-2 -7˙
˙
11 7 ˙˚
È1 2 1 ˘ È 9 0 0 ˘
– Í0 1 -1˙ + Í0 9 0 ˙ = O
Í
˙ Í
˙
ÍÎ3 -1 1 ˙˚ ÎÍ0 0 9˙˚
Matrices 5.37
È1 2 x ˘
43. (3A)¢ = Í2 1 2 ˙
Í
˙
ÍÎ2 -2 y ˙˚
Now, 9I = (3A) (3A)¢
t
t +1 t -1
t
t+2
D = t +1
t -1 t + 2
t
Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get
t t +1 t -1
2t + 1 t + 1 t - 1
1
1
3
=
0
-1
3
D=
-2
2
-2
0
2
-2
È 1 2 2 ˘ È1 2 x ˘
= Í2 1 -2 ˙ Í2 1 2 ˙
Í
˙Í
˙
ÍÎ x 2 y ˙˚ ÎÍ2 -2 y ˙˚
È
9
0
x + 4 + 2y ˘
Í
˙
0
9
2 x + 2 - 2 y˙
= Í
Í x + 4 + 2 y 2 x + 2 - 2 y x 2 + 4 + y2 ˙
Î
˚
fi x + 2y + 4 = 0
2x – 2y + 2 = 0
x2 + y2 + 4 = 9
fi x = –2, y = –1
44. Solving x + 2y = 3, 3x + 4y = 7, we get
x = 1, y = 1
\ a + 1 = 3 fi a = 2.
45. If p π –2,
3
-1
then z =
p
+2
2
1
and y =
corresponding to this value we can
p+2
get x.
If p = – 1/2, we get z = 2.
Corresponding to which we get infinite number of
value of x and y.
For p = –2, the system of equations is inconsistent.
46. The system of equations will have a unique solution
if
1 1 1
1 0
0
2 1 -1 π 0 fi 2 -1 -3 π 0
3 2 k
3 -1 k - 3
[using C1 Æ C1 – C1, C3 Æ C3 – C1]
fi (k – 3) (–1) –3 π 0 fi k π 0.
47. | A | π 0 fi 4(x + 1) – (2x – 3) (x + 2) π 0.
fi 2x2 – 3x – 10 π 0
fi x π 3 ± 89
4
48. See Theory.
49. Subtracting first equation from the second and third,
we get
y + 3z = a – 1
3y + 9z = a2 – 1
\ a2 – 1 = 3(a – 1) fi a = 1 or a = 2
50. The system of equations will have a non-trivial
solution if D = 0 where
[using C1 Æ C1 + C2]
= (2t + 1) (2 – 6) = – 4(2t – 1)
Note that D = 0 for t = –1/2 i.e., for just for one
value of t.
51. Choice (d) is not true as | A + A | = | 2A | = 8| A |
52. A = iB
Ê 1 -1ˆ Ê 1 -1ˆ
= –2B
fi A2 = i2B2 = (–1) Á
Ë -1 1 ˜¯ ÁË -1 1 ˜¯
fi A4 = (–2B)2 = 4B2 = 4(2B) = 8B
fi A8 = (8B)2 = 64B2 = 64(2B) = 128B
53. We have
3+i
i ˆ
Ê 2
7 - i˜
A¢ = Á 3 - i p
Á
˜
Ë -i 7 + i
e ¯
3 - i -i ˆ
Ê 2
Á
7 - i˜ = A
fi A*= ( A¢ ) = 3 + i p
Á
˜
Ë i
7+i
e ¯
Thus A is a Hermitian matrix.
54. Statement-2 is true since |A| π 0, implies A–1 exists.
\
AX = B
fi
A– 1 (AX) = A– 1 B
fi
(A–1 A)X = A–1 B
fi
IX = A–1 B
fi
X = A–1 B
That statement-1 is false can be seen by the following example.
È0 0 1 ˘
È0 1 0 ˘
Í
˙
Let A = 0 0 0 and B = Í0 0 0 ˙ then
Í
˙
Í
˙
ÍÎ0 0 0 ˙˚
ÍÎ0 0 0 ˙˚
AB = O but neither A = O nor B = O
55. Since B commutes with I, we can use binomial
theorem to obtain
Ê nˆ
Ê nˆ
Ê nˆ
(I + B)n = I + Á ˜ B + Á ˜ B2 + º + Á ˜ Bn
Ë1 ¯
Ë 2¯
Ë n¯
Since B2 = O, we get Br = O " r ≥ 2.
Thus,
(I + B)n = I + nB
Now,
5.38
Complete Mathematics—JEE Main
È0 p ˘
A = I + B where B = Í
˙
Î0 0 ˚
Since B2 = O, we get
È1 100p ˘
A100 =I + 100B = Í
1 ˙˚
Î0
n
n –1
56. Since a0 A + a1 A
+ º + an – 1 A + an I = O,
and an π 0, we get AB = I where
an – 1
a
a
I
B = – 0 An – 1 – 1 An – 2 – –
an
an
an
fi B = A–1.
57. If |A| π 0, A is invertible and we can write AX = B
as X = A–1 B.
\ AX = B has a unique solution and hence is consistent.
Subtracting (2) from (3) and (1) and (2), we get
the system of equations as
3x + 4y + 5z = a
(4)
x+y+z= b–a
(5)
x+y+z= c–b
(6)
As a, b, c are in A.P. b – a = c – b
\ the last two equations are identical.
From (4) and (5) we obtain
x = 4b – 5a + k
y = 4a – 3b – 2k
z= k
where k is an arbitrary complex number. Thus, the
system of equations in statement-1 is consistent.
58. Suppose A is symmetric, then A¢ = A
Since X¢AY is a 1 ¥ 1, matrix,
X¢AY = (X¢AY)¢ = Y¢A¢(X¢)¢ = Y¢AX
Next, suppose X¢AY = Y¢AX for each pair of X and
Y.
È1 ˘
È0 ˘
= Í ˙ , E2 = Í ˙ .
Let E1
Î0 ˚
Î1 ˚
\ Statement-2 is true
Also, BX = X fi (I – B)X = O
fi X = (I – B)–1 O = O
Thus, both the statements are true but statement-2
is not a correct reason for it.
Level 2
Ê a -bˆ Ê 1 - tan q
61. Á
=
Ë b a ˜¯ ÁË tan q 1
fi
1
=
abc
a +1
b
c
a
b +1
c
a
b
c +1
Using C1 Æ C1 + C2 + C3, we get
b
c
1
1
(a + b + c + 1) 1 b + 1
c
| A| =
abc
b
c +1
1
Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get
1 b c
1
| A| =
(a + b + c + 1) 0 1 0
abc
0 0 1
1
(a + b + c + 1) π 0
abc
63. A2 = 5A – 7I
=
fi A4 = (A2)2 = (5A – 7I ) (5A – 7I )
= 25 A2 – 70A + 49I
= 25 (5A – 7I ) – 70A + 49I
a12 = a21
Thus, A is symmetric.
59. For truth of statement-2, see theory
As det (Adj A) = (det A)n – 1 = O if det A = O
Thus, statement-1 is true and statement-2 is correct
reason for it.
60. We have
(I + B + B2 + … + Bk) (I – B)
= (I – B) + (B – B2) + (B2 – B3) + …
+ (Bk – Bk + 1)
k +1
=I – B
fi
- sin q cos q ˆ
˜
cos2 q ¯
Ê cos 2q - sin 2q ˆ
=Á
Ë sin 2q cos 2q ˜¯
fi a = cos2q, b = sin2q
1 + 1/ a
1
1
62.
1
1 +1 / b
1
1
1
1 + 1/ c
Taking X = E1 and Y = E2, then
E1¢AE2 = E2¢AE1
2
ˆ Ê cos q
˜¯ Á
Ë sin q cos q
I + B + B2 + … + Bk = (I – Bk + 1) (I – B)–1
= 55A – 126I
8
fi A = (A4)2 = (55A – 126I ) (55A – 126I )
= 3025 A2 – 13860A + 15876 I
= 3025 (5A – 7I) – 13860A + 158766I
= 1265A – 5299 I
Thus, a = 1265.
1
cos ( b - a ) cos (g - a )
1
cos (g - b )
64. |A| = cos (a - b )
cos (a - g ) cos ( b - g )
1
Matrices 5.39
cos a
= cos b
cos g
sin a
sin b
cos g
0 cos a
0 cos b
0 cos g
sin a
sin b
sin g
We have
0
0 =0
0
a3 + b3 + c3 – 3abc
= (a + b + c) (a2 + b2 + c2 – bc – ca – ab)
=1
fi A is singular.
fi a3 + b3 +
Èa
69. Let A = Íb
Í
ÍÎ c
65. | A(q)| = sin2q – i2 cos2q = 1.
Ê sin q
\ A(q) – 1 = Á
Ë - i cos q
- i cos q ˆ
sin q ˜¯
= A(p – q )
n
66. A B = A
n–1
c3 = 7
b c˘
d e˙
˙
e f ˙˚
Trace of A2 = (a2 + b2 + c2) + (b2 + d2 + e2)
(AB) = An – 1 (BA)
+ (c2 + e2 + f 2)
= An – 2 (AB) A
fi a2 + 2b2 + 2c2 + d2 + 2e2 + f 2 = 0
= An – 2 (BA) A = An – 2 BA2
As a, b, c, d, e, f are real numbers, we get
Continuing in this way, we get
a = b = c = d = e = f = 0.
AnB = BAn
\A=0
Similarly, ABn = BnA, (AB)n = AnBn
70. Let k = 2015. We are given
and AnBn = BnAn
(AB)k = Ak Bk and (AB)k + 1 = Ak
However, AnB = BnA does not hold.
+1
Bk
+1
We have
67. AX = l X
Ak + 1 B
k+1
= (AB)k + 1 = (AB)k (AB) = (AkBk) (AB)
fi (A – l I) X = 0
fi Ak + 1 Bk + 1 = Ak (Bk A) B
As X π 0, |A – l I | = 0
1- l
2
1
0
1
l
1 =0
fi
3
1
1- l
As A, B are non-singular matrices, we get
fi (1 – l)
ABk = Bk A.
Similarly, using (AB)k + 1 = Ak + 1 Bk + 1
and (AB)k + 2 = Ak + 2 Bk + 2
1- l -1
2
+3
1
1- l
1- l
we get AB k + 1 = B k + 1 A
1
=0
-1
Now, AB k + 1 = B k + 1 A = B(B k A)
= (1 – l) [(1 – l)2 + 1] + 3[– 2 – (1 – l)] = 0
= (1 – l)3 – 2(1 – l) – 6 = 0
fi
fi l3 – 3l2 + 5l + 7 = 0
(1)
If l1, l2, l3 are roots of (1), then
l1 + l2 + l3 = 3.
Èa b c ˘ Èa b c ˘ Èa
68. A = ÍÍb c a ˙˙ ÍÍb c a ˙˙ = Í b
Í
ÍÎ c a b ˙˚ ÍÎ c a b ˙˚ ÍÎ b
where a = a2 + b2 + c2,
2
b = bc + ca + ab.
2
As A = I, we get
a2 + b2 + c2 = a = 1
bc + ca + ab = 0
Now, (a + b + c)2 = a + 2b = 1
fia+b+c=1
= B(AB k)
b b˘
a b˙
˙
b a ˙˚
(AB) B k = (BA) B k
As B is invertible, we get AB = BA.
71. |adj A| = |A|2 = 100 = |A|2 fi |A| = ± 10
È1 ˘
72. M Í0 ˙ = M
Í ˙
ÍÎ0 ˙˚
Ê È 1 ˘ È0 ˘ˆ
Á Í-1˙ + Í1 ˙˜
Á Í ˙ Í ˙˜
Ë ÍÎ 0 ˙˚ ÍÎ0 ˙˚¯
È-1˘ È 1 ˘ È0 ˘
= Í 2 ˙ + Í 1 ˙ = Í3 ˙
Í ˙ Í ˙ Í ˙
ÍÎ 3 ˙˚ ÍÎ-1˙˚ ÍÎ2 ˙˚
È0 ˘
Ê È1˘ È1 ˘ È0 ˘ˆ
Í
˙
and M 0 = M Á Í1˙ - Í0 ˙ - Í1 ˙˜
Í ˙
Á Í ˙ Í ˙ Í ˙˜
Ë ÍÎ1˙˚ ÍÎ0 ˙˚ ÍÎ0 ˙˚¯
ÍÎ1 ˙˚
È 0 ˘ È0 ˘ È-1˘ È 1 ˘
= Í 0 ˙ - Í3˙ - Í 2 ˙ = Í-5˙
Í ˙ Í ˙ Í ˙ Í ˙
ÍÎ12 ˙˚ ÍÎ2 ˙˚ ÍÎ 3 ˙˚ ÍÎ 7 ˙˚
5.40
Complete Mathematics—JEE Main
Ê 0 0 -1ˆ Ê 0 0 -1ˆ
2. We have A = Á 0 -1 0 ˜ Á 0 -1 0 ˜ = I
Á
˜Á
˜
Ë -1 0 0 ¯ Ë -1 0 0 ¯
È1 ˘ È 0 ˘
M ÍÍ0˙˙ = ÍÍ3˙˙ fi m11 = 0
ÍÎ0˙˚ ÍÎ0˙˚
2
È0˘ È-1˘
Similarly, M ÍÍ1˙˙ = ÍÍ 2 ˙˙ fi m22 = 2
ÍÎ0˙˚ ÍÎ 3 ˙˚
3. As B is inverse of A, AB = I
\
Ê 1 -1 1 ˆ Ê 4 2 2 ˆ
Á 2 1 -3˜ Á -5 0 a ˜ = 10 I
Á
˜Á
˜
Ë 1 1 1 ¯ Ë 1 -2 3 ¯
fi
Ê10 0 5 - a ˆ Ê10 0 0ˆ
Á 0 10 a - 5˜ = Á 0 10 0˜
Á
˜ Á
˜
Ë 0 0 5 + a ¯ Ë 0 0 10¯
and m33 = 7
Thus, m11 + m22 + m33 = 9
73. (A–1 + B –1) [A – A (A + B)–1 A]
= B –1 (B + A)A–1 A[I – (A + B)–1 A]
–1
–1
= B (A + B) (A + B) [A + B – A]
fi
a=5
2
4. A – A + I = O fi I = A – A2 = A(I – A)
fi A–1 = I – A
= B –1 I B = I
74. A – B – A (A + B)–1 A + B (A + B)–1 B
= A[I – (A + B)–1 A] – [I – B (A + B)–1] B
–1
= A(A + B) [A + B – A]
– [A + B – B] (A + B)–1 B
5. For n = 1, all (a), (b), (c) and (d) hold good. We
have
È1 0 ˘ È1
A2 = Í
˙Í
Î1 1 ˚ Î1
If (a) holds for some n
0˘ È1 0˘
=
1 ˙˚ ÍÎ 2 1 ˙˚
≥ 1, then
= A (A + B)–1 B – A (A + B)–1 B
An + 1 = AnA = [nA + (n – 1)I] A
=O
= nA2 + (n – 1)A
È1 0 ˘
È1 0˘
= nÍ
- (n - 1) Í
˙
˙
Î2 1˚
Î1 1 ˚
2n - 1
0 ˘
= ÈÍ
˙
Î 3n - 1 2n - 1˚
75. A + B = AB
fi I – (A + B) + AB = I
fi (I – A) (I – B) = I
fi (I – A)–1 = I – B
fi I – (B + A) + BA = I
È1 0˘
È1 0˘
π (n + 1) Í
+ nÍ
˙
˙ = (n + 1) A + nI
Î1 1 ˚
Î0 1˚
fi A + B = BA
Similarly (b) does not hold for n + 1.
Thus, AB = BA
For (c),
fi (I – B) (I – A) = I
An +1 = AnA = [nA – (n – 1)I ]A
76. Use
= nA2 – (n – 1)A
O = A3 = (a + d)2 A
Ê xy
2
77. A = Á 0
Á
ÁË y 2
0 - x2 ˆ
0
0 ˜
˜
0 - xy ˜¯
È1 0 ˘
È1 0˘
= nÍ
- (n - 1) Í
˙
˙
Î2 1˚
Î1 1˚
1
0˘
= È
Í n + 1 1˙ = (n + 1) A - nI
Î
˚
Thus, (c) holds.
A3 = O
Previous Years' AIEEE/JEE Main Questions
2
Èa b ˘ Èa b ˘ Èa + b
1. A2 = Í
˙Í
˙=Í
Î b a ˚ Î b a ˚ Î 2ab
\ a = a2 + b2, b = 2ab
2
2ab ˘
˙
a 2 + b2 ˚
Alternatively
È0 0˘
2
Put B = Í
˙ and note that A = I + B and B =
1
0
Î
˚
O. As I and B commute.
Matrices 5.41
An = (I + B)n = I + nB
[∵ B2 = O]
= n(B + I) – (n – I) I = nA – (n – I)I
Now, A2 = I fi |A2| = |I|
A2 – B2 = (A – B)(A + B)
6.
= A2 – BA + AB – B2
fi
BA = AB
Ê a 0ˆ
commutes with
7. Each matrix of the form Á
Ë 0 a ˜¯
A.
8.
|A| = 25a fi |A2| = |A|2 = 25
fi (25a)2 = 25
fi
|A|2 = 1 fi |A| = ± 1.
Suppose |A| = 1.
In this case, A2 = I fi A = A–1
Ê a b ˆ Ê d -b ˆ
fiÁ
=
Ë c a ˜¯ ÁË -c a ˜¯
fi
1
fi
a =
5
Ê a bˆ
9. Let A = Á
Ë c d ˜¯
a = d, b = 0, c = 0
Tr(A) = 0 fi a + d = 0
fi
2a = 0 fi a = 0
Ê 0 0ˆ
In this case A = Á
Ë 0 0˜¯
Thus, in this case, if we assume statement-2 is true
then we get statement-2 is false.
Now, A2 = I fi det(A2) = 1
fi (det A)2 = 1 fi det A = ± 1.
Also, A2 = I fi A = A–1
In case |A| = –1, then
Ê a bˆ
Ê d -b ˆ
fi Á
= det A Á
˜
Ëc d¯
Ë -c a ˜¯
If det A = 1, then
A = – A–1
a = d, b = – b, c = – c fi a = d, b = c = 0.
Ê a 0ˆ
In this case A = Á
Ë 0 a ˜¯
2
|A| = 1 fi a = 1 fi a = ± 1.
Ê a b ˆ Ê d -b ˆ
fiÁ
=
fia+d =0
Ë c d ˜¯ ÁË -c a ˜¯
\ |A| = –1 fi Tr(A) = 0.
Therefore Statement-1 is true and Statement-2 is
false.
\ A = I or A = –I. A contradiction.
Thus, det (A) = –1.
Ê a bˆ
Ê d -b ˆ Ê - d
= -1Á
=
\Á
˜
Ëc d¯
Ë -c a ˜¯ ÁË c
a b˘
11. Let A = ÈÍ
˙
Îc d ˚
bˆ
- a ˜¯
\ a = –d fi Tr(A) = a + d = 0
\ Statement-1 is true and Statement-2 is false.
Ê a bˆ
10. Let A = Á
Ë c d ˜¯
Ê d -b ˆ
adj A = Á
Ë -c a ˜¯
|adj A| = ad – bc = |A|
\ Statement-2 is true.
Ê a bˆ
=A
adj (adj A) A = Á
Ë c d ˜¯
Statement-1 is also true but Statement-2 is not a
correct reason for it.
12. The matrix
È1
Íc
Í
ÍÎ e
a
1
f
b˘
d˙
˙
1 ˙˚
where exactly one of a, b, c, d, e, f is 1 and rest
of them are zeros, is invertible.
There are six such matrices.
È1 0
Also, the matrix Í 0 1
Í
ÍÎ 1 0
Thus, there are at least
invertible.
1˘
0 ˙ is invertible.
˙
0 ˙˚
7 such matrices which are
13. Adding the first two equations and subtracting the
third from the sum, we obtain
(x1 + 2x2 + x3) + (2x1 + 3x2 + x3) – (3x1 + 5x2 +
2x3) = 3 + 3 – 1
fi0=5
5.42
Complete Mathematics—JEE Main
Thus, the system of equation has no solution.
14. The system of equations will have a non-zero solution if and only if
4
k
2
k 2
4 1 =0
2 1
fi4
k 1
k 4
4 1
-k
+2
=0
2 1
2 1
2 2
19. We have
Ê 1 ˆ Ê 0ˆ Ê 1 ˆ
A(u1 + u2 ) = Au1 + Au2 = Á 0˜ + Á 1˜ = Á 1˜
Á ˜ Á ˜ Á ˜
Ë 0¯ Ë 0¯ Ë 0¯
We solve the above equation for u1 + u2.
We consider the augmented matrix
fi 4(2) – k(k – 2) + 2(2k – 8) = 0
fi k2 – 6k + 8 = 0 fi k = 2, 4
Ê 1 0 0 1ˆ
( A / B) = Á 2 1 0 1˜
Á
˜
Ë 3 2 1 0¯
Applying R3 Æ R3 – 2R2 + R1 and R2 Æ R2 – 2R1,
we get
Ê 1 0 0 1ˆ
Ê 1ˆ
( A / B) ∼ Á 0 1 0 -1˜ fi u1 + u2 = Á -1˜
Á
˜
Á ˜
Ë 0 0 1 -1¯
Ë -1¯
15. H = wI fi H70 = w70I
But w70 = (w3)23w = w
\ H70 = wI = H
16. The system of equations will have only trivial solution if and only if
1 -k
k 3
3 1
¤ (1)
fi (1) (0) – a (4 – 6) + (3)(4 – 6) = 16
fi 2a – 6 = 16 fi a = 11
1
-k π 0
-1
21. As the system of equations has no solution
3 -k
k -k
k 3
+k
+ (1)
π0
1 -1
3 -1
3 1
fi 2(k2 + k – 6)
20. |P| = |A|2 = 16
0fik
–3, 2
8
4k
k +1
=
π
k
k + 3 3k - 1
k +1
8
=
fi k 2 + 4 k + 3 = 8k
k
k +3
fi (k – 1) (k – 3) = 0 fi k = 1, 3
But
17. We have A¢ = A and B¢ = B
Now, (A(BA))¢ = (BA)¢ A¢ = (A¢B¢)A¢
= (AB)A = A(BA)
fi A(BA) is symmetric.
Similarly, (AB) A is a symmetric matrix
Also, (AB)¢ = B¢A¢ = BA
Note that AB is symmetric if and only if AB = BA.
Thus, both Statement-1 and Statement-2 are true
but Statement-2 is not a correct explanation for the
statement-1.
18. P3 = Q3 and P2Q = Q2P gives
2
8
4k
=
k + 3 3k - 1
8
8 4
4k
3
= = and
=
k +3 6 3
3k - 1 2
Thus, there is exactly one such value of k.
and for k = 3,
[∵ matrix multiplication is associative]
3
For k = 1,
3
22. Adding first two equations and subtracting from
third, we get
(a – 8)x3 = b – 15
8, we get value of x3, and substituting in
If a
first two we get values of x1 and x2. When a 8,
we get the system has a unique solution.
Thus, a = 8 and hence b = 15.
2
P –PQ=Q –QP
2
fi P (P – Q) = – Q2(P – Q)
fi (P2 + Q2) (P – Q) = 0
If det (P2 + Q2) 0, then P2 + Q2 is invertible and
hence P = Q. Therefore, det (P2 + Q2) = 0.
When a = 8 and b = 15, we solve first two equations in terms of x3 to obtain infinite number of
solutions.
23. 3p + 3q + 2r = 3
(i)
4p + 2q = 0
(ii)
Matrices 5.43
p + 3q + 2r = 1
(iii)
From (i) and (iii) 2p = 2 or p = 1
When p = 1, q = –2 [from (ii)]
a bˆ
27. Let A = ÊÁ
Ë c d ˜¯
A2 = I fi det (A)2 = 1
fi det (A) = ± 1.
Now, 2p + q – r
Now, A2 = I fi A–1 = A
1
= 2 p + q - [1 - p - 3q ]
2
1
= 2(1) + ( -2) - [1 - 1 - 6] = -3
2
24. As the system of equation has a unique solution.
Ê d - bˆ Ê a b ˆ
fi det( A) Á
=
Ë -c a ˜¯ ÁË c d ˜¯
(1)
If det (A) = 1, then
a = d, –b = b, –c = c
1 a 0
D= 0 1 a π0
a 0 1
fi a = d, b = 0, c = 0.
Using C1 Æ C1 + C2 + C3, we get
Thus, in this case A = I or A = –I.
\ det (A) = 1 fi a2 = 1 fi a = ± 1.
But A π I, –I.
1 a 0
D = (1 + a ) 1 1 a
1 0 1
Therefore det(A) = –1.
fi Statement-2 is true.
Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get
Now, from (1)
1
a
0
D = (1 + a ) 0 1 - a a
0 -a 1
Ê d -b ˆ Ê a b ˆ
( -1) Á
=
Ë -c a ˜¯ ÁË c d ˜¯
fi a + d = 0 fi tr(A) = 0
= (1 + a) (1 – a + a2) = 1 + a3
D
0fia
Thus, Statement-1 is true and Statement-2 is correct
reason for it.
–1 [∵ a Œ R]
a bˆ
25. Let A = ÊÁ
; a, b, c Π{0, 1, 2}.
Ë c a ˜¯
BB¢ = (A–1A¢ )(A–1A¢ )¢
= (A–1A¢ )[(A¢ )¢(A–1)¢]
= A–1(A¢A)(A–1)¢
|A| = a2 – bc
When a = 0, bc
28.
0 fi b, c Π{1, 2}
There are four such values of b, c, viz. (b, c) =
(1, 1), (1, 2), (2, 1) and (2, 2)
When a = 1, both b and c cannot be 1. In this case
there are 8 such pairs of (b, c).
When a = 2, both b and c cannot be 2. In this case
also, there are 8 such pairs of (b, c).
2˘ È 1
2˘ È 9 - 4˘
È1
26. A2 = Í
˙
Í
˙=Í
˙
Î 4 - 3˚ Î 4 - 3˚ Î - 8 17 ˚
2
A + 4A – 5I
8˘ È - 5 0 ˘
È 9 - 4˘ È 4
=Í
+Í
˙
˙+Í
˙
Î - 8 17 ˚ Î16 - 12 ˚ Î 0 -5˚
È8 4 ˘
È2 1˘
=Í
= 4Í
˙
˙
Î8 0 ˚
Î 2 0˚
= A–1(AA¢ ) (A¢ )–1
= (A–1A)[A¢(A¢)–1]
= (I)(I) = I
29. As B2 = 0, Br = 0 " r ≥ 2.
Now, (I + B)50 = I +
50
C50 B50
50
C 1B +
= I + 50B
fi (I + B)50 – 50B = I
fi det(I + B)50 – 50B = 1
30. Applying C2 ´ C3, we get
È1 3 2˘ È0 1 0˘
A Í0 3 2˙ = Í1 0 0˙
Í
˙ Í
˙
ÍÎ 0 1 1 ˙˚ ÍÎ 0 0 1 ˙˚
50
C 2B 2 + +
[∵ Br = 0 " r ≥ 2]
5.44
Complete Mathematics—JEE Main
Applying C1 ´ C2, we get
È3 1 2 ˘ È1
A Í3 0 2 ˙ = Í 0
Í
˙ Í
ÍÎ1 0 1 ˙˚ ÍÎ 0
È3 1
-1
fi A = Í3 0
Í
ÎÍ1 0
È6 ˘
31. AB = Í ˙
Î8 ˚
fi
0 0˘
1 0˙
˙
0 1 ˙˚
2˘
2˙
˙
1 ˙˚
Now, A4 – I = O = A2 + I
and A3 – I = –A – I = –A + A2 = A(A – I)
and A3 + I = –A + I = A4 – A = A(A3 – I)
But A2 + I = O and A(A2 – I) = – 2A.
\ A2 + I
A(A2 – I).
36. We know A(Adj A) = |A|I2
\ AA¢ = |A|I2
y + 2x + x = 6
3y – x + 2 = 8
fi
fi A3 = –A and A4 = I
3x + y = 6
–x + 3y = 6
fi
3x + y = –x + 3y
fi
y = 2x
32. Let C = AB – BA, then
C¢ = (AB – BA)¢ = (AB)¢ – (BA)¢
= B¢A¢ – A¢B¢
= (–B)A – A(–B)
= –BA + AB = C
fi AB – BA is symmetric.
33. AAT = 9I
È 1 2 2 ˘ È1 2 a ˘
È1 0 0˘
Í
˙
Í
˙
fi 2 1 -2 2 1 2 = 9 Í 0 1 0 ˙
Í
˙Í
˙
Í
˙
ÍÎ a 2 b ˙˚ ÎÍ 2 -2 b ˙˚
ÍÎ 0 0 1 ˙˚
È
9
0
a + 4 + 2b ˘ È 9 0 0 ˘
Í
˙
0
9
2a + 2 - 2b ˙ = Í 0 9 0 ˙
fiÍ
Í
˙
Í a + 4 + 2b 2a + 2 - 2b a 2 + 4 + b 2 ˙ ÍÎ 0 0 9 ˙˚
Î
˚
È5a -b ˘ È 5a 3˘
È1 0 ˘
fi Í
= (10a + 3b) Í
Í
˙
˙
˙
Î 3 2 ˚ Î -b 2 ˚
Î0 1 ˚
2
2
È1 0 ˘
È
˘
fi Í25a + b 15a - 2b ˙ = (10a + 3b) Í
˙
Î0 1 ˚
13 ˚
Î 15a - 2b
fi 25a2 + b2 = 10a + 3b, 15a – 2b = 0, 10a + 3b
= 13
Now, b = (15/2)a 100 +
45
a = 13
2
fi a = 2/5, b = 3
\ 5a + b = 2 + 3 = 5
Ê cos a
37. P = Á
Ë - sin a
Ê cos a
P¢ = Á
Ë sin a
sin a ˆ
where a = p/6
cos a ˜¯
- sin a ˆ
cos a ˜¯
Ê 1 0ˆ
We have PP¢ = Á
Ë 0 1˜¯
–1
fi P¢ = P
Thus, Q = PAP–1 fi P–1QP = A
fi A2 = (P–1QP)2 = P–1Q2P
fi A3 = P–1Q3P
Continuing like this we get,
fi a + 2b = –4, a – b = –1, a2 + b2 = 5
A2015 = P–1Q2015P = P¢Q2015P
fi (a, b) = (–2, –1)
Ê 0 1ˆ
We have A = I + B where B = Á
Ë 0 0˜¯
34. |5 adj A| = 5
fi 53 |adj A| = 5
fi |adj A| = 1/25
2
fi |A| = 1/25 fi |A| = ± 1/5
È 0 -1˘ È 0 -1˘
2
35. A = Í
˙Í
˙ = -I
Î 1 0˚ Î 1 0˚
Ê 0 1ˆ Ê 0 1ˆ
Ê 0 0ˆ
Also, B2 = Á
= Á
˜
Á
˜
Ë 0 0¯ Ë 0 0¯
Ë 0 0˜¯
Ê 0 0ˆ
fi Br = Á
"r≥2
Ë 0 0˜¯
Ê 1 2015ˆ
Thus, A2015 = I + 2015B = Á
1 ˜¯
Ë0
Matrices 5.45
2016
– 2A
=A
2014
2
(A – 2A – I)
=A
2014
(A – ( 2 + 1)I)(A + ( 2 – 1)I)
38. Let B = A
2015
–A
2014
Note that B2 = O.
A2 = (I + B)2 = I2 + 2IB + B2
= I + 2B
-4 -1
We have |A| =
= –1
3 1
|A – ( 2 + 1)I| =
3
A = AA2 = (I + B)(I + 2B)
= I2 + BI + 2(IB) + 2B2
-4 - ( 2 + 1)
-1
3
1 - ( 2 + 1)
= I + 3B
Thus,
[∵
B2 = O]
2A2 – A3 = 2(I + 2B) – (I + 3B)
= I + B = A.
= (–4 – ( 2 + 1)(1 – ( 2 + 1) + 3
3. Statement-2 is true.
= ( 2 + 1)2 + 3( 2 + 1) – 1 = 5 + 5 2
|A + ( 2 – 1)I| =
-4 + ( 2 - 1)
-1
3
1 + ( 2 - 1)
= (–4 + ( 2 – 1))(1 + ( 2 – 1)) + 3
2
= ( 2 – 1) – 3( 2 – 1) – 1
tr(AB – BA) = tr(AB) – tr(BA) = 0
But tr(I) = 2.
\
AB – BA
I.
Thus, both statements are true and Statement-2 is
a correct explanation for Statement-1.
4. As A РaI is invertible for all a ΠR.
=5–5 2
Thus det (B) = (–1)2014 (5 + 5 2 )(5 – 5 2 )
= –25
39. A2 – 5A = 7I = O
fi 7I = A(5I – A) = (5I – A)A
det (A – aI)
" a ΠR.
0
0
" a ΠR.
fi a2 – (a + d)a + ad – bc
0 " a ΠR.
fi (a – a)(d – a) – bc
Therefore
I = AB = BA
1
where B = (5I – A)
7
\ A–1 = B =
[See Theory]
(a + d)2 – 4(ad – bc) < 0
(a – d)2 + 4bc < 0
fi
1
(5I –A)
7
Therefore, bc < 0.
fi Statement-1 is true.
Also, a2 + d2 – 2ad + 4bc < 0
We have
fi
0 £ a2 + d2 < 2ad – 4bc
= A(5A – 7I) – 2(5A – 7I) – 3A + I
fi
bc <
= 5A2 – 7A – 10A + 14I – 3A + I
1
Thus, bc < min ÊÁ 0, ad ˆ˜
Ë 2 ¯
A3 – 2A2 – 3A + I
= 5(5A – 7I) – 20A + 15I
= 25I – 35I – 20A + 15I
1
ad .
2
5. Statement-2 is false as
= 5(A – 4I)
adj(AB) = (adj B) (adj A)
Thus, Statement-2 is also true.
(adj A) (adj B)
Previous Years' B-Architecture Entrance
Examination Questions
1. adj(AB) = (adj B) (adj A)
2. A = I + B
È0 1 0˘
where B = Í 0 0 0 ˙
Í
˙
ÍÎ 0 0 0 ˙˚
[See Theory]
But Statement-1 is true as
A(adj A) = |A|I2.
[See Theory]
6. det (A) = ad – bc
Note that det (A) can take value –1, 0 or 1. We
have
det(A) = 1 ¤ ad = 1, bc = 0
5.46
Complete Mathematics—JEE Main
¤ a = 1, d = 1 or (b = 0, c = 0, b = 0; c = 1;
b = 1, c = 0)
and det(A) = –1 ¤ ad = 0 or bc = 1
This is also possible in 3 cases.
Ê 1 -k ˆ
-1
Also, Sk = ÁË
0 1 ˜¯
Ê 1 2 nˆ Ê 1 - k ˆ
\ ( S2 ) n ( Sk ) -1 = Á
Ë 0 1 ˜¯ ÁË 0 1 ˜¯
Ê 1 2n - k ˆ
= S2n - k
= Á
Ë0
1 ˜¯
\ A–1 exists in 6 cases.
a b ˘ Èa b ˘
7. A2 = ÈÍ
˙Í
˙
Îc d ˚ Îc d ˚
È0 1˘
9. Let A = I + C where C = Í
˙
Î0 0˚
È a 2 + bc b(a + d ) ˘
=Í
˙
ÍÎ c(a + d ) bc + d 2 ˙˚
AS C2 = O, we get Cr = 0 " r > 2.
Now,
A10 = (I + C)10
È a 2 + bc
2b ˘
=Í
˙
ÍÎ 2c
bc + d 2 ˙˚
Èa 2 - 2a + bc
˘
0
\ A - 2A = Í
˙
ÍÎ
0
a 2 - 2d + bc ˙˚
= I + 10C +
10
C2 (C2) + … +
10
È1 10˘
= I + 10C + O = Í
˙
Î0 1 ˚
2
Since A¢ = A2 – 2A, we get
È 1 -10 ˘
fi Adj( A10 ) = Í
˙
Î0 1 ˚
2
˘
0
È a c ˘ È a - 2a + bc
=
˙
Íb d ˙ Í
2
˚ ÍÎ
Î
0
a - 2d + bc ˙˚
b
As È 1
Íb
Î 3
b2 ˘
= 10 A10 + adj( A10 ) , we get
b4 ˙˚
fi b = 0, c = 0, a = a2 – 2a, d = d2 – 2d
È b1
fi Í
Îb3
b2 ˘ È10 100 ˘ È 1 -10 ˘
+
=
b4 ˙˚ ÍÎ 0 10 ˙˚ ÍÎ 0 1 ˙˚
fi a(a – 3) = 0 and d(d – 3) = 0
fi a = 0 or a = 3 and d = 0 or d = 3
But then a + d cannot be 2.
\
fi
È11 90 ˘
= Í
˙
Î 0 11 ˚
b1 = b4 = 11, b2 = 90, b3 = 0,
\ b1 + b2 + b3 + b4 = 112
S=f
Ê 1 2ˆ
= I + B where
8. S2 = Á
Ë 0 1˜¯
Ê 1 0ˆ
Ê 0 2ˆ
I =Á
and B = Á
˜
Ë 0 1¯
Ë 0 0˜¯
2
Note that B = O, therefore, Br = O " r ≥ 2.
n
C10(C10)
n
Thus, (S2) = (I + B) = I + nB
È 1 2n ˘
=Í
˙
Î0 1 ˚
10. |adj A| = |A|2 = 36
1 -2 4
fi 4 1 1 = 36
-1 k 0
fi (–1)
-2 4
1 4
–k
= 36
1 1
4 1
[Expanding along R3]
fi (–1)(–2 – 4) – k(1 – 16) = 36
fi 6 + 15k = 36 fi k = 2
CHAPTER SIX
FUNDAMENTAL PRINCIPLES OF
COUNTING
The Sum Rule
Suppose that A and B are two disjoint events (mutually exclusive); that is, they never occur together. Further suppose
that A occurs in m ways and B in n ways. Then A or B can
occur in m + n ways. This rule can also be applied to more
than two mutually exclusive events.
The Product Rule
Suppose that an event X can be decomposed into two stages,
A and B. Let stage A occur in m ways and suppose that these
stages are unrelated, in the sense that stage B occurs in n
ways regardless of the outcome of stage A. Then event X
occurs in mn ways. This rule is applicable even if event X
can be decomposed in more than two stages.
PERMUTATIONS AND COMBINATIONS
Suppose we have a set of n distinct objects {x1, x2, , xn}
from which we have to take a sample of r (1 £ r £ n) objects.
In how many ways can this be done? However, the problem
is not completely specified; in particular, the phrase ‘take a
sample’ is somewhat ambiguous.
1. Is the order in which we take the sample important?
For example, if r = 2, do we consider first picking
x1 and then x2 to be different from picking x2 and
then x1?
2. Do we allow repetition of objects? For example, do
we allow a sample to consists of two x1?
Since each question has two answers, the product rule
says that our original problem is really made up of four
separate problems.
1. How many such samples are there if the order is
important and we allow repetitions?
2. How many such samples are there if the order is
important and we do not allow repetitions?
3. How many such samples are there if the order is
not important and we allow repetitions?
4. How many such samples are there if the order is
not important and we do not allow repetitions?
We now introduce some terminology. If the order is important, then we say that our sample consists of arrangements. If the order is not important then we have selections.
If repetition is allowed, then we are using replacement.
An arrangement of r objects with replacement is called an
r-sequence (or simply sequence). Vehicle numbers are examples of sequences.
An r-permutation (or simply permutation) is an arrangement without replacement. The different orderings of a deck
of cards are examples of permutations. A selection of r objects with replacement is called an r-multiset. The letters
used to form the word MATHEMATICS can be thought of
as a multiset with two Ms, two As, two Ts, one H, one E, one
I, one C and one S. An r-combination (or simply combination) is a selection of r objects without replacement.
In Table 6.1 we illustrate the four different problems and
their solutions. We take two objects from {x1, x2, x3}.
Table 6.1
Repetition
x1 x1
x1 x2
x1 x3
x2 x1
x2 x2
r-sequence
x2 x3
x3 x1
x3 x2
x3 x3
{x1, x1}
{x1, x2}
{x1, x3}
{x2, x2}
r-multiset
{x2, x3}
{x3, x3}
No repetition
x1 x2
x1 x3
x2 x1
x2 x3
r-permutation
x3 x1
x3 x2
{x1, x2}
{x1, x3}
{x2, x3}
r-combination
6.2
Complete Mathematics—JEE Main
What happens when r = 0? We may view the r-combination as a subset of r elements, and thus a 0-combination
may be viewed as the empty set. Since there is exactly one
empty set, we may say that there is exactly one 0-combination. Mathematicians also say that there is exactly one 0-sequence, one 0-permutation, and one 0-multiset. By relaxing
the definitions we may use the terms sequence and permutation interchangeably and call a multiset, a combination.
Remark
1. Note that even if one object is lying on the circle, then the
circle has to be treated as a row, so far as arranging the
objects around the circle is concerned.
2. Note that if no distinction is to be made between
clockwise and counter-clockwise arrangements, then the
number of arrangements is equal to (n – 1)!/2.
3. The number of ways of arranging r identical objects and
SOME IMPORTANT RESULTS
1. The number of sequences (that is, permutations) of n
distinct objects taken r at a time, when repetition of objects
is allowed, is nr (r > 0).
2. The number of permutations of n distinct objects taken r
(0 £ r £ n) at a time, when repetition is not allowed, is given
by
n!
n
Pr =
(n - r )!
(n – r) distinct objects along a circle is
Illustration
L
L
By definition Cr = 0 if r > n.
5. The number of ways of selecting zero or more objects
out of p1 identical objects of first kind, p2 identical objects of second kind, …, pr identical objects of rth kind is
(p1 +1) (p2 +1) … (pr +1).
6. The number of combinations of n distinct objects taken
r (£ n) at a time, when k (0 £ k £ r) particular objects
always occur, is n – kCr – k.
7. The number of combinations of n distinct objects taken
r at a time, when k (1 £ k £ n) never occur, is n – kCr.
8. If n > 1, nP0 < nP1 < nP2 < . . . < nPn – 1 = nPr
9. If n = 2m +1, then
n
C0 < nC1 < nC2 … < nCm = nCm+1
n
Cm+1 > nCm+2 > nCm+3 > … > nCn
10. If n = 2m, then
n
C 0 < nC 1 < nC 2 … < nC m
n
Cm > nCm+1 > nCm+2 … > nCn
CIRCULAR PERMUTATION
The number of ways of arranging n distinct objects around
a circle is (n – 1)!.
M
M
L
M
M
M
M
L
L
n!
p1 ! p2 ! pr !
n
1
In how many ways can 8 ladies and 5 men can be seated
around a round table so that no two men are together.
Solution: 8 ladies can be arranged arround a round table in (8 –
1)! ways. After the ladies have been arranged, there are 8 places
for men. We can place 5 men at these places in 8P5 ways.
Thus, required number of ways is (7!) (8P5).
Conventionally, nPr = 0 for r > n.
3. The number of permutations of n objects taken all toge
ther, when p1 of the objects are alike and of one kind, p2 of
them are alike and of the second kind, , pr of them are alike
and of the rth kind, where p1 + p2 +
+ pr = n is given by
4. The number of combinations of n distinct objects taken r
(0 £ r £ n) at a time is given by
n!
Ê nˆ
n
ÁË r ˜¯ = Cr = (n - r )! r !
(n - 1)!
r!
L
M
M
L
L
SOME IMPORTANT IDENTITIES
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
n
C 0 = 1 = nC n.
n
C r = nC n – r
(0 £ r £ n).
n
C r – 1 + nC r = n + 1C r.
n
Cr = nCs implies r = s or r + s = n.
n - r +1 n
n
Cr =
Cr – 1
(1 £ r £ n).
r
If n is even, then the greatest value of nCr is nCm,
where m = n/2. If n is odd, then the greatest value
of nCr is nCm, where m = (n – 1)/2 or (n + 1)/2.
n
C 0 + nC 1 + nC 2 +
+ nC n = 2 n.
n
C 0 + nC 2 +
= nC 1 + nC 3 +
= 2 n – 1.
2n + 1
2n + 1
2n + 1
C0 +
C1 +
+
Cn = 22n
2n + 1
Cn + 1 + 2n + 1Cn + 2 + º + 2n+1C2n + 1 = 2 2n
DIVISION OF IDENTICAL OBJECTS
The number of ways of distributing n identical objects
among r persons giving zero or more to each is equal to
the number of ways of arranging n identical object of one
kind and (r – 1) identical objects of second kind (separators)
in a row
(n + r - 1)! n + r – 1
=
=
Cr – 1
n!(r - 1)!
Permutations and Combinations 6.3
2
Illustration
Illustration
To distribute 10 identical toys among 3 children, we may use
two separators as follows:
1st child
2nd child
3rd child
4
Find the number of ways of distributing 10 identical toys
among 3 children so as give ai (1 £ i £ 3) to the ith children,
where 2 £ a1 £ 4, 3 £ a2 £ 5, 1 £ a3 £ 7.
The required number of ways
= coefficient of x10 in (x2 + x3 + x4) (x3 + x4 + x5) (x + . . . + x7)
= coefficient of x10 in x6 (1 + x + x2)2 (1 + ... + x6)
2
1st child
2nd child
1st child
2nd child
3rd child
3rd child
etc.
The required number of ways
=
(10 + 2)! 12
= C2 = 66.
10! 2!
Alternative method
1. The number of distributing n identical objects among r
persions giving zero or more to each
= coefficient of xn in (1 – x) – r
= coefficient of xn in (1 + x + x2 + . . .)r
2. The number of ways of distributing n identical object
among r persons giving at least one to each.
= the number of ways of distributing remaining (n – r) identical objects (after giving one to each of r persons) among r
persons giving zero or more to each
= (n – r) + (r – 1)Cr – 1 = n – 1Cr – 1
Illustration
3
(i). To distribute 10 toys among 3 children giving at
least one to each, we first give one toy to each
of the child and distribute the remaining 7 toys
among 3 children giving zero or more to each in
7+3–1
C3 – 1 = 9C2 = 36 ways.
(ii). However if we wish to give at least two
to first, at least three to second and at
least one to the third, then, we first give
(2 + 3 + 1) toys to the three children as desired
and distribute the remaining 4 toys among three
children giving zero or more to each. This can be
done in 4 + 3 – 1C3 – 1 = 6C2 = 15 ways.
(iii). If wish to distribute n identical object among r persons so that kth person (1 £ k £ r) does not get more
than ak, then the desired number of ways
= Coeffcient of xn in (1 + x + . . . + xa1) (1 + x + . . . + xa2)
. . . (1 + x + . . . + xar)
Ê 1 - x3 ˆ Ê 1 - x 7 ˆ
= coefficient of x in Á
Ë 1 - x ˜¯ ÁË 1 - x ˜¯
4
= coefficient of x4 in (1 – 2x3) (1 – x)– 3 [ignore any power
that is more than 4]
= coeffcient of x4 in (1 – 2x3) (1 + 3C1 x + 4C2 x2 + 5C3 x3 +
6
C4 x4 + . . .)
6
= C4 – 2(3C1) = 15 – 6 = 9
Finding Number Integral Solutions of a Linear
Equation
The number of non-negative integral solution of
x1 + x2 + .... + xr = n,
where n Œ N » {0}, and r Œ N
= the number of ways of distributing n
identical objects among r persons giving
zero or more to each
= n + r – 1C r – 1.
MULTIPLICATION OF TWO INFINITE
SERIES
The following informal method of multiplying two infinite
power series is also useful:
(a0+ a1x + a2x2 + a3x3 +…) (b0 + b1 x+b2x2+…)
= c0 + c1x + c2x2 + c3x3 +…
where cn= a0bn + a1bn–1 + a2bn–2 +… anb0 for all n ≥ 0.
DISTRIBUTION INTO GROUPS
1. The number of ways in which n distinct objects can be
split into three groups containing respectively r, s and t objects (where r, s and t are distinct and r + s + t = n) is
( nC r) ( n – rC s) ( n – r – sC t)
=
n!
(n - r )!
(n - r - s)!
r !(n - r )! (n - r - s)! s ! (n - r - s - t )! t !
=
n!
r ! s! t !
2. Suppose mn distinct objects are to be divided into m
groups, each containing n objects and the order of the
groups is not important. Then the number of ways of doing
this is given by
(m n)!
m !(n !)m
6.4
Complete Mathematics—JEE Main
If, however, the order of groups is important, then the
number of ways is given by
(m n)!
(n !)m
SELECTION
The last integer amongst 1, 2, , [n/p] which is divisible
by p is
Èn ˘
È [ n / p] ˘
Í p ˙ p=Í 2˙ p
Î
˚
Îp ˚
Since the remaining integers are not divisible by p, we
get
1. The number of ways of selecting one or more objects out
of n distinct objects is 2n – 1.
2. Suppose we have n distinct objects, and p like objects of
one kind, q like objects of a second kind, r like objects of a
third kind, etc. Then the number of ways of selecting one or
more objects from these objects is
2n (p + 1) (q + 1) (r + 1)
– 1.
Continuing in this way, we get
EXPONENT OF PRIME p IN n!
where ps £ n < ps + 1.
Let Ep(m) denote the exponent of the prime p in the prime
factorization of positive integer m. We have
Ep(n!) = Ep (1 ◊ 2 ◊ 3 ◊ 4
(n – 1) ◊ n)
The last integer amongst 1, 2, 3, , (n – 1), n which is
divisible by p is [n/p]p, where [x] denotes greatest integer
£ x.
Therefore,
Ê
Èn˘ ˆ
Ep(n!) = Ep Á p ◊ 2 p ◊ 3 p Í ˙ p˜
Ë
Î p˚ ¯
because the remaining integers from the set {1, 2, 3,
(n – 1), n} are not divisible by p.
Ê
Èn˘ ˆ
Ep(n!) = Ep Á p ◊ 2 p ◊ 3 p Í ˙ p˜
Ë
Î p˚ ¯
Èn˘ È n ˘
Ep(n!) = Í ˙ + Í 2 ˙ + E p
Î p˚ Î p ˚
Ê
È n ˘ˆ
ÁË 1◊ 2 Í p2 ˙˜¯
Î ˚
Èn˘
Èn˘ È n ˘
Ep(n!) = Í ˙ + Í 2 ˙ + + Í s ˙
Î p˚ Î p ˚
Îp ˚
5
Illustration
Find exponent of 3 in 32!
As 3 divides 3, 6, 9...., 30 amongst 1, 2, . . . 31, 32,
E3(32!) = E3(3 ¥ 6 ¥ 9 ¥ . . . ¥ 30)
= E3(310 ¥ 1 ¥ 2 ¥ . . . ¥ 10)
= 10 + E3(3 ¥ 6 ¥ 9)
[Ignore 1, 2, 4, 5, 7, 8, 10 as 3does not divide these.]
= 10 + E3(33 ¥ 1 ¥ 2 ¥ 3)
,
= 10 + 3 + E3(3)
= 10 + 3 + 1 = 14.
È 31 ˘
È 31 ˘
È 31 ˘
and 1 = Í 3 ˙
Note, 10 = Í ˙ , 3 =
ÍÎ 32 ˙˚
Î3˚
Î3 ˚
Èn˘
Ê
È n ˘ˆ
= Í ˙ + E p Á 1◊ 2 ◊ 3 Í ˙˜
Ë
Î p˚
Î p ˚¯
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Example 1: If
n+5
1
pn +1 = (11)(n - 1)
2
equal to:
(a) 10, 11
(b) 9, 10
(c) 8, 7
(d) 6, 7
Ans: (d).
n+5
1
Pn +1
Solution:
(11) (n – 1) =
n+ 3
2
Pn
= (n + 5)! ¥ 3!
(n + 3)!
4!
( n+3 pn ) then n is
fi
fi
fi
1
1
(11) (n – 1) = (n + 5) (n + 4)
2
4
2
22n – 22 = n + 9n + 20
n2 – 13n + 42 = 0 fi n = 6, 7
Example 2: If n + 2C3 = n + 3P2 – 20, then n is equal to:
(a) 6
(b) 5
(c) 4
(d) 3
Ans (d).
1
Solution:
(n + 2) (n + 1)n = (n + 3) (n + 2) – 20
6
Permutations and Combinations 6.5
fi
n3 + 3n2 + 2n = 6(n2 + 5n + 6) – 120
fi
n3 – 3n2 – 28n + 84 = 0
fi
(n – 3) (n2 – 28) = 0
As n is a positive integer, we get n = 3
Example 3: If 10Cr – 1 > 2(10Cr) then r is equal to
(a) 6, 7, 8, 9
(b) 7, 8, 9
(c) 8, 9, 10
(d) 5, 6, 9, 10
Ans: (c).
10
Solution: Note that 0 £ r £ 10. Now
Cr -1
10
Cr
TIP
You can directly reach the answer by substituting the given
choices in (1).
>2
fi
10!
r !(10 - r )!
>2
.
(r - 1)! (10 - r + 1)!
10!
fi
r
> 2 fi r > 22 – 2r
11 - r
fi
3r > 22 fi r ≥ 8 fi r = 8, 9, 10.
Example 6: If there are 62 onto mapping from a set
X containing n elements to the set Y = {– 1, 1}, then n is
equal to:
(a) 4
(b) 5
(c) 6
(d) 7
Ans: (c)
1 n
1
1
( Pr + 1) = (nPr) = (nPr – 1), then
a
b
c
2
b – (a + b)c is equal to.
(a) 0
(b) 1
(c) – 1
(d) – 2
Ans: (a).
Example 4: If
n
Solution:
Pr + 1
n
Pr
=
a
a
(n - r )!
fi
=
b
(n - r - 1)! b
n–r=
Similarly
n–r+1=
\
fi
b
c
b a
- =1
c b
b2 – ac = bc fi b2 – (a + b)c = 0.
Exemple 5: Let Qn be the number of possible quadrilaterals formed by joining vertices of an n sided regular
polygon. If Qn + 1 – Qn = 20, then value of n is:
(a) 8
(b) 7
(c) 6
(d) 5
Ans: (c).
Solution: Qn = Number of ways of choosing four vertices out of n
= nC 4
\
20 = Qn + 1 – Qn = n + 1C4 – nC4
= nC 4 + nC 3 – nC 4 = nC 3
[ nC r – 1 + nC r = n + 1C r]
fi
1
n (n – 1) (n – 2) = 20
6
Solution: Let X = {x1, x2, . . . , xn}.
Each xi can have two images viz – 1 and 1. Thus, there are
2n mappings from X to Y. But there are exactly two mapping which are not onto. These are when all the elements
are mapped to – 1 or when all the elements are mapped to 1.
\ there are 2n – 2 onto mapping from X to Y.
Set
2n – 2 = 62 = fi 2n = 64 = 26
fi
n=6
Example 7: The maximum possible number of points
of intersection of 7 straight lines and 5 circles is:
(a) 111
(b) 109
(c) 107
(d) 105
Ans: (a)
a
b
fi
n3 – 3n2 + 2n – 120 = 0
n3 – 6n2 + 3n2 – 18n + 20n – 120 = 0
(n – 6) (n2 + 3n + 20) = 0
As n ΠN, n = 6.
fi
fi
fi
(1)
Solution: Maximum possible number of points of intersection of 7 lines is 7C2 = 21.
Maximum possible number of points of intersection of 5
circles in 2 (5C2) = 20
[ two circles can intersect in two distinct points.]
Maximum possible number of points of intersection of 5
circles and 7 lines is 2(5) (7) = 70
Thus, maximum possible number of points of intersection is
21 + 20 + 70 = 111.
Example 8: The number of injective functions from a
set X containing m elements to a set Y containing n elements,
for m > n is:
(b) (m – n)!
(a) mPn
(d) 0
(c) mCn
Ans. (d)
Solution:
TIP
If X and Y are two finite sets and f : X Æ Y is an injective
6.6
Complete Mathematics—JEE Main
function from X to Y then n ≥ m where m = n(X) and n = n(Y).
The number of injective functions from X to Y is
n(n – 1) (n – 2) . . . (n – m + 1) = nPm
As m > n, there does not exist any injective function from X
to Y.
Example 9: If nPr = 2520 and nCr = 21 then n is equal to:
(a) 6
(b) 7
(c) 8
(d) 10
Ans: (b)
Solution: (r!) (nCr) = nPr fi (r!) (21) = 2520
fi
r! = 120 fi r = 5
7!
1
n
Now
C5 = 21 =
(7 ¥ 6) =
5
! 2!
2!
fi
(a) 2
(c) 4
Ans: (b)
(b) 3
(d) 5
Solution: Note that 0 £ 3r £ 15
15!
(3r )!(15 - 3r )! 3
=
¥
fi 0 £ r £ 5. Also,
(14 - r )! (r + 1)!
15!
11
(3r )! (15 - 3r )! 3
=
(14 - r )! (r + 1)! 11
fi
For the denominator, to be divisible by 11, 11|(14 – r)!.
Setting 14 – r = 11.
fi
r=3
Putting r = 3, in LHS of (1) we get
9! 6 !
6¥ 5
3
=
=
= RHS of (1)
11! 4! 11 ¥ 10 11
LHS of (1)
n=7
Example 10: If 15Cr + 1 : 15C3 r = 3 : 11, then value of r is:
(1)
\
r=3
LEVEL 1
Straight Objective Type Questions
Example 11: If nC4, nC5 and nC6 are in A.P., then a value
of n can be
(a) 6
(c) 8
Ans. (b)
(b) 7
(d) 9
\
Solution: As nC4, nC5 and nC6 are in A.P.,
2 ( nC 5) = nC 4 + nC 6
n
fi
2=
n
C4
C5
+
n
C6
n
C5
fi
n
C6
n
C7
5
n-5
+
n-4
6
2=
fi
n = 7, 14.
fi
n > 13
Solution: We have
fi
n2 – 21n + 98 = 0
n
Cr -1
n
fi
Cr
n–1
(b) 15
(d) 28
Solution: Since mCr–1 + mCr = m +1Cr we can write the
=
36
n!
r ! (n - r )!
3
fi
=
84
(r - 1)! (n - r + 1)!
n!
7
r
3
=
n - r +1 7
n
Example 12: The least positive integer n for which
given inequality as
n
C 6 < nC 7
7
<1
n-6
Example 13: If nCr–1 = 36, nCr = 84 and nCr+1 = 126,
then the value of r is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (c)
Similarly,
C5 + n–1C6 < nC7 is
(a) 14
(c) 16
Ans. (a)
fi
the least value of n is 14.
n!
5!(n - 5)!
n!
5!(n - 5)!
+
=
4 ! (n - 4)!
n!
6 ! (n - 6)!
n!
fi
<1
n
Cr
Cr +1
=
84
126
fi
5r + 3 = 2n
Solving we obtain r = 3.
fi
10 r = 3n +3
fi
r +1 2
=
n-r 3
fi
5r + 3 = 2n
Example 14: In a group of 8 girls, two girls are sisters.
The number of ways in which the girls can sit in a row so
that two sisters are not sitting together is
(a) 4820
(b) 1410
(c) 2830
(d) 30240
Ans. (d)
Permutations and Combinations 6.7
Solution: The required number of ways
= the number of ways in which 8 girls can
sit in a row – the number of ways in which
two sisters sit together
= 8! – (2) (7!) = 30240.
Example 18: The number of subsets of the set A = {a1,
a2, …, an} which contain even number of elements is
(b) 2n –1
(a) 2n–1
n –2
(d) 2n
(c) 2
Ans. (a)
Example 15: The number of words that can be formed
by using the letters of the word MATHEMATICS that start
as well as end with T is:
(a) 80720
(b) 90720
(c) 20860
(d) 37528
Ans. (b)
Solution: For each of the first (n – 1) elements a1, a2,
…, an–1 we have two choices: either ai (1 £ i £ n–1) lies in
the subset or ai doesn’t lie in the subset. For the last element
we have just one choice. If even number of elements have
already been selected, we do not include an in the subset,
otherwise (when odd number of elements have been selected), we include it in the subset.
Thus, the number of subsets of A = {a1, a2,…, an} which
contain even number of elements is equal to 2n–1.
Solution: The word MATHEMATICS contains 11 letters viz. M, M, A, A T, T, H, E, I, C, S. The number of words
that begin with T and end with T is
9!
= 90720.
2! 2!
n
r =0
(a)
equals:
r
n
a
4 n
(d) (n–1)an
n
Solution: Let bn =
n
2bn = Â
r
r + (n - r )
r =0
bn =
n
r
 nC
r =0
fi
r =0
(b)
(c) nan
Ans. (a)
r
 nC
, then
r
n
a
2 n
fi
n
1
 nC
Example 16: If an =
n
Cr
n-r
=Â
n
r =0
Cn - r
n
= nÂ
r =0
1
n
Cr
n
=Â
n-r
r =0
n
Cr
= nan
n
a .
2 n
Example 17: m men and w women are to be seated in
a row so that no two women sit together. If m > w, then the
number of ways in which they can be seated is:
(a)
(c)
Ans. (a)
m ! (m + 1)!
(m - w + 1)!
m+w
Cm (m – w)!
(b)
m
Cm–w (m – w)!
(d) none of these
Solution: We first arrange the m men. This can be done
in m! ways. After m men have taken their seats, the women
must choose w seats out of (m + 1) seats marked with X
below.
X M X M X M X …X M X
1st 2nd 3rd
mth (m + 1)th
m+1
They can choose w seats in Cw ways and arrange w women in w! ways.
Thus, the required number of arrangements is
m ! (m + 1)! w ! m ! (m + 1)!
=
m! (m +1Cw) (w!) =
w ! (m + 1 - w)! (m + 1 - w)!
Example 19: The number of positive integers
< 1,00,000 which contain exactly one 2, one 5 and one 7 in
its decimal representation is
(a) 2940
(b) 7350
(c) 2157
(d) 1582
Ans. (a)
Solution: We may consider a number of up to 5-digits
to be a number of the form
XXXXX
where X is a digit from 0 to 9. The digit 2 can occupy any of the
five places, 3 can occupy any of the remaining 4 places and 7
any of the 3-remaining places. The remaining 2 places can be
filled up by 7 digits. Thus, there are (5) (4) (3) (7) (7) = 2940
positive integers in the desired category.
Example 20: The number of ways of factoring 91,000
into two factors, m and n, such that m > 1, n > 1 and
gcd (m, n) = 1 is
(a) 7
(b) 15
(c) 32
(d) 37
Ans. (a)
Solution: We have 91,000 = 23 × 53 × 7 × 13
Let A = {23, 53, 7, 13} be the set associated with the prime
factorization of 91,000. For m, n to be relatively prime, each
element of A must appear either in the prime factorization of
m or in the prime factorization of n but not in both. Moreover, the 2 prime factorizations must be composed exclusively from the elements of A. Therefore, the number of
relatively prime pairs m, n is equal to the number of ways
of partitioning A into 2 unordered non-empty subsets. We
can partition A as follows:
{23}»{53, 7, 13}, {53} » {23, 7, 13}
{7}»{23, 53, 13}, {13} » {23, 53, 7}
and
{23, 53}»{7, 13},{23, 7} » {53, 13},
{23, 13}» {53, 7}
Therefore, the required number of ways = 4 + 3 = 7.
6.8
Complete Mathematics—JEE Main
Example 21: The sum S =
Solution: We have
31!
1
1
=
=
E = 31
= 2–36 = (23)–12 = 8–12
2 (32 !) 231 (32) 236
1
1
1
1
1
+
+
+
+
9! 3! 7! 5! 5! 7! 3! 9!
equals
(a)
29
10 !
(b)
Thus, x = –12.
Example 25: The value of
17
17
17
(1 + 17) ÊÁ 1 + ˆ˜ ÊÁ 1 + ˆ˜ … ÊÁ 1 + ˆ˜
Ë
2 ¯Ë
3 ¯ Ë 19 ¯
E=
is
19
19
19
(1 + 19) ÊÁ 1 + ˆ˜ ÊÁ 1 + ˆ˜ … ÊÁ 1 + ˆ˜
Ë
2¯Ë
3 ¯ Ë 17 ¯
210
(d)
7!
11
(c)
210
8!
2
9!
Ans. (a)
Solution: We can write S as follows
1 10
S=
[ C1 + 10C3 + 10C5 + 10C7 + 10C9]
10!
1
(29 ).
=
10!
Example 22: Let n = 2015. The least positive integer
k for which
k(n2) (n2 – 12) (n2 – 22) (n3 – 32) … (n2 – (n – 1)2) = r!
for some positive integer r is
(a) 2014
(b) 2013
(c) 1
(d) 2
Ans. (d)
Solution: We can rewrite the given expression as
k(n2) (n – 1) (n + 1) (n – 2) (n + 2) (n – 3) (n + 3)…
(n + n – 1) (n – n + 1) = r!
fi kn (1) (2)… (n – 1) n(n + 1) (n + 2)… (2n – 1) = r!
fi
kn (2n – 1)! = r!
\
To convert L.H.S. to a factorial, we shall require,
k = 2 which will convert it into (2n)!
Example 23: If 0 < r < s £ n and nPr = nPs , then value
of r + s is
(a) 2n – 2
(b) 2n – 1
(c) 2
(d) 1
Ans. (b)
n!
n!
=
Solution: nPr = nPs fi
(n - r )! (n - s)!
fi
(n – r)! = (n – s)!
As r < s, n – r > n – s. But the only two different factorials
which are equal are 0! and 1!. Thus n – r = 1 and n – s = 0
fi r = n–1 and s = n.
fi
of x is
(a) – 7
(c) – 10
Ans. (d)
(c) 2/19
Ans. (a)
(b)
36
C17
(d)
36
C18
Solution: We have
Ê kˆ Ê kˆ Ê kˆ
(1 + k ) Á 1 + ˜ Á 1 + ˜ … Á 1 + ˜
Ë 2¯ Ë 3¯ Ë n¯
=
(1 + k )(2 + k )(3 + k ) … (n + k ) (n + k )!
=
(2)(3)… (n)
k ! n!
= n+k Ck
Thus, both the numerator and the denominator of E equals
36
C17 = 36C19.
\
E = 1.
Example 26: Kunal Gaba has n objects, each of weight
w. He weighs them in pairs and finds the sum of the weights
of all possible pairs is 120g. When his friend Rakshit weighs
them in triplets, the sum of all possible weights is 240g. The
value of n is
(a) 7
(b) 6
(c) 5
(d) 10
Ans. (b)
Solution: According to the given condition
(nC2) (2w) = 120 fi n(n – 1)w = 120
and
(nC3) (3w) = 240 fi n(n –1) (n – 2)w = 480
Thus,
fi
n(n - 1)(n - 2)w 480
=
n(n - 1)w
120
n–2=4
fi
n=6
Example 27: If [y] denote the greatest integer £ y,
r + s = 2n – 1.
Example 24: If E =
(a) 1
1 2 3 4 30 31
. . . .
= 8x, then value
4 6 8 10 62 64
(b) – 9
(d) – 12
2
Èx˘
Èx˘
and 2 Í ˙ + 3 Í ˙ = 20, then x lies in the smallest interval
Î8 ˚
Î8 ˚
[a, b) where b– a is equal to
(a) 6
(b) 5
(c) 4
(d) 8
Ans. (d)
2
Èx˘
Èx˘
Solution: 2 Í ˙ + 3 Í ˙ - 20 = 0
Î8 ˚
Î8 ˚
Permutations and Combinations 6.9
5
Èx˘
or
fi Í ˙ =
Î8 ˚
2
-4
Èx˘
As Í ˙ is an integer, we take
Î8 ˚
Èx˘
=–4
ÎÍ 8 ˚˙
fi
– 4 £ x/8 < – 3
fi – 32 £ x < – 24. Thus, a = – 32, and b = – 24. Therefore,
b–a=8
m
Ê 10ˆ Ê 20 ˆ
Ê pˆ
Example 28: The sum  Á ˜ Á
˜¯ (where ÁË q ˜¯ = 0
Ë
¯
Ë
i
m
i
i =0
if p < q) is maximum where m is
(a) 5
(b) 10
(c) 15
(d) 20
Ans. (c)
m
Solution: We have
Ê 10ˆ Ê 20 ˆ
 ÁË i ˜¯ ÁË m - i˜¯
i =0
= the number of ways of choosing m persons
out of 10 men and 20 women
= the number of ways of choosing m persons
out of 30 persons
= 30Cm
But 30Cm is maximum for m = 15.
Example 29: Let Tn denote the number of triangles
which can be formed by using the vertices of a regular polygon of n sides. If Tn +1 – Tn = 21, then n equals
(a) 5
(c) 6
Ans. (b)
(b) 7
(d) 4
Solution: The number of triangles that can be formed
by using the vertices of a regular polygon is nC3. That is,
T n = nC 3
Now,
Tn +1 – Tn = 21
n +1
C3 – nC3 = 21
fi
fi nC2 + nC3 – nC3 = 21
[ n+1Cr= nCr–1 + nCr]
fi
1
n(n – 1) = 21
2
fi
n = – 6 or 7.
As n is a positive integer, n = 7.
Example 30: An eight digit number divisible by 9 is to
be formed by using 8 digits out of the digits 0, 1, 2, 3, 4, 5,
6, 7, 8, 9 without replacement. The number of ways in which
this can be done is
(a) 9!
(b) 2(7!)
(c) 4(7!)
(d) (36) (7!)
Ans. (d)
Solution: We have 0 + 1 + 2 + 3 … + 8 + 9 = 45
To obtain an eight digit number exactly divisible by 9, we
must not use either (0, 9) or (1, 8) or (2, 7) or (3, 6) or
(4, 5). [Sum of the remaining eight digits is 36 which is
exactly divisible by 9.]
When, we do not use (0, 9), then the number of required
8 digit numbers is 8!.
When one of (1, 8) or (2, 7) or (3, 6) or (4, 5) is not used,
the remaining digits can be arranged in 8!– 7! ways.{0 cannot be at extreme left.}
Hence, there are 8! + 4(8! – 7!) = (36) (7!) numbers in the
desired category.
Example 31: The number of rational numbers lying in
the interval (2015, 2016) all whose digits after the decimal
point are non-zero and are in decreasing order is
9
(a)
 9 Pi
i=1
9
(c) 2 –1
Ans. (c)
10
(b)
 9 Pi
i=1
(d) 210–1
Solution: A rational number of the desired category is
of the form 2015. x1 x2 … xk where 1 £ k £ 9 and 9 ≥ x1 >
x2 > … > xk ≥ 1. We can choose k digits out 9 in 9Ck ways
and arrange them in decreasing order in just one way. Thus,
the desired number of rational numbers is 9C1 + 9C2 + …+
9
C9 = 29–1.
Example 32: The number of functions f from the set A
= {0, 1, 2} in to the set B = {0, 1, 2, 3, 4, 5, 6, 7} such that
f(i) £ f (j) for i < j and i, j Œ A is
(a) 8C3
(b) 8C3 + 2(8C2)
10
(d) 10C4
(c) C3
Ans. (c)
Solution: A function f : A Æ B such that f (0) £ f (1) £
f (2) falls in one of the following four categories.
Case 1
f (0) < f (1) < f (2)
There are 8C3 functions in this category.
Case 2
f (0) = f (1) < f (2)
There are 8C2 functions in this category.
Case 3
f (0) < f (1) = f (2)
There are again 8C2 functions in this
category.
Case 4
f (0) = f (1) = f (2)
There are 8C1 functions in this category.
Thus, the number of desired functions is
8
C3 + 8C2 + 8C2 + 8C1 = 9C3 + 9C2 = 10C3.
Example 33: The number of positive integral solution
of the equation x1 x2 x3 x4 x5 = 1050 is
6.10
Complete Mathematics—JEE Main
(a) 1800
(c) 1400
Ans. (d)
(b) 1600
(d) none of these
Solution: Using prime factorization of 1050, we can
write the given equation as
x1 x2 x3 x4 x5 = 2 × 3 × 52 × 7
We can assign 2, 3 or 7 to any of 5 variables. We can assign
entire 52 to just one variable in 5 ways or can assign 52 =
5 × 5 to two variables in 5C2 ways. Thus, 52 can be assigned in
5
C1 + 5C2 = 5 + 10 = 15 ways
Thus, the required number of solutions is 5 × 5 × 5 × 15
= 1875.
Example 34: The number of ways in which we can
arrange the digits 1, 2, 3, …, 9 such that the product of five
digits at any of the five consecutive positions is divisible
by 7 is
(a) 7!
(b) 9P7
(c) 8!
(d) 5(7!)
Ans. (c)
Solution: Let an arrangement of 9 digit number be
x1 x2 x3 x4 x5 x6 x7 x8 x9. Note that we require product of each
of (x1, x2, x3, x4, x5); (x2, x3, x4, x5, x6); …; (x5, x6, x7, x8, x9)
is divisible by 7.
This is possible if the 5th digit is 7.
Therefore, we can arrange the 9 digits in desired number of
ways in 8! ways.
Example 35: Let X be a set containing n elements. The
number of reflexive relations that can be defined on X is
2
2
(b) 2n – n
(a) 2n
2
2
(c) n(2n – n)
(d) n(2n )
Ans. (b)
Solution: Let X = {x1, x2, . . . xn}. The set X ¥ X contains
n2 elements. A reflexive relation R on X must contain (xi, xi)
for 1 £ i £ n, i.e. R must contain the n elements (x1, x1), (x2, x2)
. . . (xn, xn) and any subset of the set containing remaining
n2 – n elements. Therefore, the number of reflexive relations
2
that can be defined on X is 2n – n
Example 36: Suppose X contains m elements and Y
contain n elements. The number of functions from X to Y is
(a) nm
(b) nm
n
(c) Pm
(d) nCm
Ans. (b)
Solution: Let X = {x1, x2, . . . xm} and
Y = {y1, y2, . . . yn}
For each xi (1 £ i £ m) we have n possible images. Therefore,
the number of functions from X to Y is nm.
Example 37: Three boys and three girls are to be seated
around a circular table. Among them the boy X does not want
any girl as a neighbour and girl Y does not want any boy as a
neighbour. The number of possible arrangement is:
(b) 4
(a) 9
(c) 8
(d) 6
Ans. (b)
Solution: Clearly X should sit between the two remaining boys and Y should sit between the remaining two girls.
We have to arrange two groups B1 X B2 and G1 Y G2
along a circle. This can be done is (2 – 1)! ways. But group
of boys (girls) can be arranged in 2! ways. Thus, the required number of ways is (2) (2) (1) = 4 ways.
Example 38: If nCr : nCr + 1 : nCr + 2 = 1 : 2 : 3, then r is
equal to
(a) 5
(b) 4
(c) 3
(d) 0
Ans. (b)
n
Cr + 1 n - r
2
Solution:
= n
=
r +1
1
Cr
and
fi
fi
3
n - r -1
=
2
r+2
3
2(r + 1) - 1
=
2
r+2
3r + 6 = 4r + 2 fi r = 4.
[
n – r = 2(r + 1)]
Example 39: The number of ways selecting three numbers from 1 to 30, so as to exclude every selection of three
consecutive number is:
(a) 4030
(b) 4031
(c) 4032
(d) 4035
Ans. (c)
Solution: The numbers of ways of selecting 3 number
out of 30 is 30C3. But this include 28 ways of selecting 3
consecutive numbers. [viz (1, 2, 3), (2, 3, 4), 3, 4, 5), ....,
(28, 29, 30)]
Thus, required number of ways is 30C3 – 28 = 4032.
Example:40: If all permutations of the letters of the
word P E N C I L are arranged as in a dictionary, then 413th
word is
(a) L I C N E P
(b) L I C N P E
(c) L I C P N E
(d) L I C P E N
Ans. (d)
Solution: Letters in the word PENCIL are C, E, I, L,
N, P.
There are 5! words each beginning with C, E and I. That is,
there are 3(5!) = 360 words beginning with C, E, I
Words beginning with L have serial numbers 361 to 480.
4! = 24 words beginning with LC have serial numbers 361
to 384. Words beginning with LE are numbered from 385
to 408. Words beginning with LIC are numbered from 409
Permutations and Combinations 6.11
to 414. Now 414th word is L I C P N E and 413th word is
L I C P E N.
Example: 41: Out of 10 consonants and 4 vowels,
words with 6 consonants and 3 vowels are formed. The
number of such words is
7
9
(b)
(12!)
(12!)
(a)
11
10
(c)
7
(11!)
10
(d)
Case 2
Case 3
4
(12!)
5
Ans. (a)
Solution: The number of ways of choosing 6 consonants out of 10 is 10C6 and 3 vowels out of 4 in 4C3. Thus
7
number of such words is (10C4) (4C3) (9P9) =
(12!)
11
Example: 42: Let L1, L2, L3 be three distinct parrallel
lines in the XY-plane, p distinct the points are taken on each
of the three lines. The maximum number of triangles than
can be formed by these 3p points is:
(b) p3 + 3(pC2)
(a) p2(4p – 3)
2
(d) (p + 1)3 – 1
(c) p (3p – 4)
Ans. (a)
L1
L2
L3
Solution: To obtain a triangle we may take one point
from each of the three lines. This will give us p3 triangles.
Alternatively, we may choose two lines and take one point
on one of them and two points on the other line. This will
give us
(3C2) (2!) (p) (pC2) = 3p2 (p – 1) triangles
\ required number of triangles is p3 + 3p2 (p – 1) = p2(4p – 3)
Alternative Solution
Choose three points not on the same lines. Number of
ways = 3pC3 – 3(pC3) = p2(4p – 3)
Example : 43: The number of ways of selecting 4 letters out of the letters of the word MINIMAL is
(a) 16
(b) 17
(c) 18
(d) 20
Ans. (b)
Solution: Letters of the word , M I N I M A L are (M,
M) (I, I), A, N, L
We can select 4 letters from letters of the word MINIMAL
as follows:
Case 1
All letters are distinct.
This can be done in 5C4 = 5 ways
Exactly two letters are identical.
We can choose two identical letters in 2C1
ways and two remaining distinct letters in
4
C2 = 6 ways
\ In this case the selection can be made
in (2) (6) = 12 ways.
Two pairs of identical letters are selected.
This can be done in 2C2 = 1 way
\ two number of ways
= 5 + 12 + 1 = 17.
Ï È18 a ˘
¸
: a, b Œ N ˝
Example 44: Let S = Ì Í
˙
Ó Î b 25˚
˛
The number of singular matrices in S is
(a) 18
(b) 25
(c) 450
(d) infinite
Ans: (a)
È18 a ˘
2
2
Solution: det Í
˙ = 2 ¥ 3 ¥ 5 – ab = 0
b
25
Î
˚
¤
ab = 2 ¥ 32 ¥ 52
¤
a | 2 ¥ 32 ¥ 52
That is a is a divisor of 2 ¥ 32 ¥ 52
Therefore, a is of the form 2a 3b 5g where
a Π{0, 1}, b, g Π{0, 1, 2}.
Thus number of elements in S is 2 ¥ 3 ¥ 3 = 18.
Example 45: The number of positive integers n such
that 2n divides n! is
(a) exactly 1
(b) exactly 2
(c) infinite
(d) none of these
Ans. (d)
Solution: The exponent of 2 in n! is given by
Èn˘ È n ˘ È n ˘
E = Í ˙ + Í 2 ˙ + Í 3 ˙ +…
Î2 ˚ Î2 ˚ Î2 ˚
where [x] denotes greatest integer £ x, As [x] £ x " x,
È n ˘
ÍÎ 2 m ˙˚ = 0 after finite number of terms. Thus we get
n
n2
n n
E < + 2 + 3 +… =
=n
2 2
1
1/ 2
2
Thus, there is no positive integer for which 2n divides n!
Example 46: The range of the function 7 – x P x – 3 is
(a) {1, 2, 3, 4}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3}
(d) {1, 2, 3, 4, 5}
Ans. (c)
Solution: We must have 7 – x ≥ 1, x – 3 ≥ 0 and
x–3£7–x
fi
x £ 6, x ≥ 3 and x £ 5
Thus,
3 £ x £ 5.
\
Range of 7 – x P x – 3 is {4P0, 3P1, 2P2} = {1, 3, 2}
6.12
Complete Mathematics—JEE Main
6
Example 47: The value of 50C4 +
(a)
(c)
Ans. (b)
56
55
C3
C4
(b)
(d)
56
55
Â
r =1
56 – r
C3 is
C4
C3
6
50
Solution:
C4 +
 56 – r C3
r =1
= 50C4 + (55C3 + 54C3 + º + 50C3)
= ( 50C4 + 50C3) + (51C3 + º + 55C3)
= (51C4 + 51C3) + ( 52C3 + º + 55C3 )
= (52C4 + 52C3) + (53C3 + º + 55C3 )
= ºº
= 56C4
Example 48: A class consists of 4 boys and g girls.
Every Sunday five students, including at least three boys go
for a picnic to Appu Ghar, a different group being sent every
week. During, the picnic, the class teacher gives each girl in
the group a doll. If the total number of dolls distributed was
85, then value of g is
(a) 15
(b) 12
(c) 8
(d) 5
Ans. (d)
Solution: Number of groups having 4 boys and 1 girl
= ( 4C 4) ( gC 1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g–1)
Thus, the number of dolls distributed
= g(1) + (2) [2g (g – 1)] = 4g2 – 3g
2
We are given 4g – 3g = 85 fi g = 5.
Example 49: Sum of the factors of 9! which are odd
and are of the form 3m + 2, where m is a natural number is
(a) 40
(b) 45
(c) 51
(d) 54
Ans. (a)
Solution: We have 9! = 27 × 34 × 5 × 7
Odd factors of the form 3m + 2 are neither multiples of 2 nor
multiples of 3. So the factors may be 1, 5, 7, 35 of which just
5 and 35 are of the form 3m + 2. Their sum is 40.
Example 50: Sum of all three digit numbers (no digit
being zero) having the property that all digits are perfect
squares, is
(a) 3108
(b) 6216
(c) 13986
(d) none of these
Ans. (c)
Solution: The non-zero perfect square digits are 1, 4 and 9.
1 can occur at units place in 3 × 3 = 9 ways.
\ Sum due to 1 at units place is 1 × 9. Similarly, sum
due to 1 at tens place is 1 × 10 × 9 and sum due to 1
at hundreds place is 1 × 100 × 9. We can deal with
the digits 4 and 9 in a similar way.
Thus, sum of the desired number is
(1 + 4 + 9) (1 + 10 + 100) (9) = 13986
Example 51: The number of ordered pairs (m, n), m, n
Œ {1, 2, …, 100} such that 7m + 7n is divisible by 5 is
(a) 1250
(b) 2000
(c) 2500
(d) 5000
Ans. (c)
Solution: Note that 7r (r ΠN) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively.)
Thus, 7m + 7n cannot end in 5 for any values of m, n ΠN.
In other words, for 7m + 7n to be divisible by 5, it should
end in 0.
For 7m + 7n to end in 0, the forms of m and n should be as
follows:
m
n
1
2
3
4
4r
4r + 1
4r + 2
4r + 3
4s + 2
4s + 3
4s
4s + 1
Thus, for a given value of m there are just 25 values of n for
which 7m + 7n ends in 0. [For instance, if m = 4r, then n =
2, 6, 10, …, 98]
\ there are 100 × 25 = 2500 ordered pairs (m, n) for
which 7m + 7n is divisible by 5.
Example 52: An n-digit number is a positive number
with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only three digits 2, 5 and 7. The
smallest value of n for which this is possible is
(a) 6
(b) 7
(c) 8
(d) 9
Ans. (b)
Solution: The number of n-digit distinct numbers that
can be formed using digits 2, 5 and 7 is 3n. We have to find
n, so that 3n ≥ 900
fi
3n–2 ≥ 100 fi
n–2≥5
fi
n ≥ 7.
Thus, the smallest value of n is 7.
Example 53: The number of ways of arranging p numbers out of 1, 2, 3, …, q so that maximum is q–2 and minimum is 2 (repetition of number is allowed) such that maximum and minimum both occur exactly once, (p > 5, q > 3) is
(b) pC2 (q –3)q–1
(a) p–3Cq–2
(c) pC2× qC3
Ans. (d)
(d) p(p–1) (q – 5)p–2
Permutations and Combinations 6.13
Solution: First we take one of the numbers as 2 and
one another as q –2. We can arrange these two numbers in
p(p–1) ways. We have to choose remaining p– 2 numbers
from the numbers 3, 4, …, q – 4, q – 3. This can be done in
(q – 5)p–2 ways.
Thus, the total number of ways of arranging the numbers
in desired way is
p(p –1) (q – 5) p–2.
Example 54: The number of ways of selecting 4 cards
of an ordinary pack of playing cards so that exactly 3 of them
are of the same denomination is
(a) 2496
(b) 13C3 ¥ 4C3 ¥ 48
52
(d) none of these
(c) C3 ¥ 48
Ans. (a)
Solution: We can choose one denomination in 13C1
ways, then 3 cards of this denomination can be chosen in 4C3 ways and one remaining card can be chosen in 48C1 ways. Thus, the total number of choices is
(13C1) (4C3) (48C1) = 13 × 4 × 48 = 2496.
Example 55: The number of integers x, y, z, w such that
x + y + z + w = 20 and x, y, z, w ≥ – 1, is
(b) 25C3
(a) 24C3
26
(c) C3
(d) 27C3
Ans. (d)
Solution: Put x = a – 1, y = b – 1, z = c – 1, w = d – 1,
then a, b, c, d ≥ 0 and (a – 1) + (b – 1) + (c – 1) + (d – 1) = 20
fi
a + b + c + d = 24
The number of non-negative integral solutions of this equation is
24 + 4 –1
C4 –1 = 27C3
Example 56: If 20% of three subsets (i.e., subsets containing exactly three elements) of the set A = {a1, a2, …, an}
contain a1, then value of n is
(a) 15
(b) 16
(c) 17
(d) 18
Ans. (a)
Solution: The number of subsets of A containing exactly three elements is nC3 whereas the number of three subsets
of A that contain a1, is n –1C2. We are given,
n–1
C2 =
20
100
( n C3 )
(n - 1)(n - 2) 1 n(n - 1) (n - 2)
fi
=
fi n = 15.
2
5
(6)
Example 57: There are three piles of identical yellow,
black and green balls and each pile contains at least 20 balls.
The number of ways of selecting 20 balls if the number of
black balls to be selected is twice the number of yellow
balls, is
(a) 6
(b) 7
(c) 8
(d) 9
Ans. (b)
Solution: Let the number of yellow balls be x, that of
black be 2x and that of green be y. Then
x + 2x + y = 20 or 3x + y = 20
fi
y = 20 – 3x.
As 0 £ y £ 20, we get 0 £ 20 – 3x £ 20
fi
0 £ 3x £ 20
\
The number of ways of selecting the balls is 7.
or
0£x£6
Example 58: The total number of permutations of
n (>1) different things taken not more that r at a time, when
a thing may be repeated any number of times, is
(a)
n
(nr - 1)
n -1
(b)
nr -1
n -1
(c)
nr + 1
n +1
(d)
nr + n
n -1
Ans. (a)
Solution: When k (1 £ k £ r) things are arranged, the
number of possible arrangements, when repetition is allowed is
n ¥ n ¥ºº¥ n = n k
k times
Thus, the total possible arrangements is
n + n2 + n3 + … + nr =
n (nr - 1)
.
n -1
Example 59: The exponent of 7 in the prime factorization of 100C50 is
(a) 0
(b) 2
(c) 4
(d) none of these
Ans. (a)
100 !
.
50 ! 50 !
È 50 ˘ È 50 ˘
The exponent of 7 in 50! is Í ˙ + Í 2 ˙ = 7 + 1 = 8,
Î 7 ˚ Î7 ˚
È100 ˘ È100 ˘
+
and the exponent of 7 in 100! is Í
Î 7 ˙˚ ÍÎ 72 ˙˚
Solution: We have 100C50 =
= 14 + 2 = 16
Thus, exponent of 7 in the prime factorization of
16 – 2(8) = 0.
100
C50 is
Example 60: In a certain test there are n questions. In
this test 2k students gave wrong answers to at least (n – k)
questions, where k = 0, 1, 2, …, n. If the total number of
wrong answers is 4095, then value of n is
6.14
Complete Mathematics—JEE Main
(a) 11
(c) 13
Ans. (b)
(b) 12
(d) 15
(a) 603
(c) 601
Ans. (c)
Solution: The number of students answering at least r
questions incorrectly is 2n– r.
\ the number of students answering exactly r (1 £ r £ n–1)
questions incorrectly is 2n–r – 2n – (r +1).
Also, the number of students answering all questions
wrongly is 20 = 1.
Thus, the total number of wrong answers is
1(2n–1 – 2n –2) + 2(2n – 2 – 2n –3) + 3(2n – 3 – 2n– 4) + …
+ (n– 1) (21– 20)+ n (20)
= 2n–1 + 2n–2 + … + 20 = 2n – 1.
Now, 2n–1 = 4095 fi 2n = 4096 = 212 fi n = 12.
Example 61: The number of ways of permuting letters
of the word ENDEANOEL so that none of the letters D, L,
N occurs in the last five positions is
(a) 5!
(b) 2(5!)
(c) 7(5!)
(d) 21(5!)
Ans. (b)
Solution: Letters of the word ENDEANOEL are
3E’s, 2N’s, 1D, 1O, 1L, 1A
Letters D, L, N can be permuted at first 4 places in
4!
ways
2!
5!
ways.
and the remaining letters can be permuted
3!
\ required number of ways 4! ¥ 5! = 2(5!)
2 ! 3!
n
Example 62: Sum of the series
 (r 2 + 1) (r !)
is
r =1
(a) (n + 1)!
(c) n(n + 1)!
Ans. (c)
(b) (n + 2)! – 1
(d) none of these
Solution: We can write
r2 + 1= (r + 2) (r + 1) – 3 (r + 1) + 2
Thus,
n
n
r =1
r =1
 (r 2 + 1) (r !) =  [(r + 2) (r + 1) - (r + 1)
-2 {(r + 1) - 1}] r !
n
=
n
 [(r + 2)! - (r + 1)!] - 2 {(r + 1)!- r !}
r =1
r =1
= (n + 2)! – 2! – 2{(n + 1)! – 1} = n (n + 1)!
Example 63: If letters of the word SACHIN are arranged in all possible ways and are written out as in a dictionary, then the word SACHIN appears at serial number
(b) 602
(d) 600
Solution: Letters appearing in the word SACHIN are
A, C, H, I, N, S
The words beginning with letters A, C, H, I and N
appear before the word SACHIN.
There are 5(5!) = 600 words beginning with A, C, H, I
and N.
Word SACHIN appears is the first word beginning with
S. Therefore, SACHIN appears at serial number 601.
Example 64: The set S = {1, 2, 3, º 12} is to be partitioned into three sets A, B, C of equal size. Thus A » B »
C = S, A « B = B « C = C « A = f. The number of ways to
partition S is
12 !
12 !
(b)
(a)
3
3! (3!)3
3! ( 4 !)
(c)
12 !
)3
(4!
(d)
12 !
(3!)4
Ans. (a)
Solution: Each of the three sets A, B, C contains exactly
4 elements.
Thus, the number of ways of partitioning the set S is
12 !
1 12
1 12 !
8!
(1) =
¥
C4 )( 8 C4 )( 4 C4 ) =
(
3! ( 4 !)3
3!
3! 4 ! 8! 4 ! 4!
Example 65: Ten letters of an alphabet are given. Words
with five letters are formed by these given letters. Then the
number of words which have at least one letter repeated is
(a) 69760
(b) 30240
(c) 99748
(d) none of these
Ans. (a)
Solution: Number of five letter words that can be
formed from 10 letters
= 10 ¥ 10 ¥ 10 ¥ 10 ¥ 10 = 105
Number of five letter words that have none of their letter
repeated
= 10P5 = 10 ¥ 9 ¥ 8 ¥ 7 ¥ 6 = 30240
\ Number of words which have at least one letter repeated
= 105 – 30240 = 69760
Example 66: The number of ways in which six ‘+’ and
four ‘–’ signs can be arranged in a line such that no two ‘–’
signs occur together is
(a) 30
(b) 35
(c) 6!5!
(d) 10!
Ans. (b)
Permutations and Combinations 6.15
Solution: Six ‘+’ signs can be arranged in just one way.
There are seven places for ‘–’ signs as shown in the following figure marked with X.
X+X+X+X+X+X+X
We can choose 4 places out of 7 in 7C4 = 35 ways.
Example 67: The remainder when x = 1! + 2! + 3! + 4!
+ …+ 100! is divided by 240, is
(a) 153
(b) 33
(c) 73
(d) 187
Ans. (a)
Solution: For r ≥ 6; r! is divisible by 240. Thus, when x
is divided by 240, the remainder is 1! + 2! + … + 5! = 153.
Example 68: If x Œ N and x –1C4 – x – 1C3 <
then x can take
(a) 8 values
(c) 6 values
Ans. (b)
5 x– 2
( P2),
4
(b) 7 values
(d) none of these
Solution: For 1 £ x £ 3, each of x –1C4, x – 1C3 and x – 2P2
is zero. For x = 4, x –1C4 = 0, x –1C3 = 1 and x–2P2 = 2 and the
inequality is valid trivally. For x ≥ 5, we have
x -1
C4 x -1 C3 5
<
x -2
P2 x - 2 P2 4
fi
( x - 1)( x - 4) x - 1 5
<
4
4!
3!
fi
x2 – 9x – 22 < 0 fi (x – 11) (x + 2) < 0
fi
x – 11 < 0 [ x + 2 > 0]
Thus, 5 £ x < 11. Hence, x can take 7 values viz. 4, 5, 6, 7,
8, 9 and 10.
Example 69: The total number of ways in which 5 balls
of different colours can be distributed among three persons
so that each person gets at least one ball is:
(a) 75
(b) 150
(c) 210
(d) 243
Ans. (b)
Solution: 5 balls can be distributed to 3 persons by
giving (2, 2, 1) balls or by giving (3, 1, 1) balls. Each of the
above distribution has three such ways. Thus, the required
number of ways
= (3) (5C2) (3C2) (1C1) + 3(5C3) (2C1) (1C1)
= (3) (10) (3) (1) + (3) (10) (2) (1) = 150
Example 70: Let n ≥ 2 be an integer. Take n distinct
points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent
points by blue and rest by red. If the number of red and blue
line segments are equal than value of n is
(a) 4
(c) 6
Ans. (b)
(b) 5
(d) 7
Solution: Total number of line segments is nC2. The
number of line segments joining adjacent points is n and
number of lines joining non-adjacent points in nC2 – n. We
are given
n
fi
C2 – n = n fi
1
n(n – 1) = 2n
2
n = 5.
Example 71: The number of seven digit integers, with
sum of the digits equal to 10 and formed by using the digits
1, 2 and 3 only, is
(a) 55
(b) 66
(c) 77
(d) 88
Ans. (c)
Solution: Let xi (1 £ i £ 7) be the digit at the ith place.
As 1 £ xi £ 3, and x1 + x2 + …+ x7 = 10, at most one xi can
be 3.
Two cases arise.
Case 1.
Exactly one of xi’s is 3. In this case exactly one of the remaining xi’s is 2.
In this case, the number of seven digit numbers is
7!
= 7 ¥ 6 = 42
5!
Case 2.
None of xi’s is 3.
In this case exactly three of xi’s is 2 and the remaining four
xi’s are 1.
In this case, the number of seven digit number is
7!
= 35
3! 4 !
Hence, the required seven digit numbers is 42 + 35 = 77.
Example 72: Let S = {1, 2, 3, 4}. The total number of
unordered pairs of disjoint subsets of S is equal to
(a) 25
(b) 34
(c) 42
(d) 41
Ans. (d)
Solution: Let A and B be two subsets of S. If x ΠS, then
x will not belong to A « B if x belongs to at most one of A,
B. This can happen in 3 ways.
Thus, there are 34 = 81 subsets of S for which A « B = f.
Out of these there is just one way for which A = B = f.
As, we, are interested in unordered pairs of disjoint sets, the
number of such subsets is
1 4
(3 - 1) + 1 = 41.
2
6.16
Complete Mathematics—JEE Main
Example 73: How many different words can be formed
by jumbling the letters in the word MISSISSIPPI in which
no two S are adjacent?
6
7
8
(b) (6 ) (7) ( C4 )
(a) (8) ( C4 ) ( C4 )
7
(c) (6 ) (8) ( C4 )
Ans. (d)
6
8
(d) (7) ( C4 ) ( C4 )
Solution: We can permute M, I, I, I, I, P, P in
Solution: This can be done in the following ways:
(i) paint the central triangle with one of the three colours;
(ii) paint each of the remaining triangles with any one
of the two remaining colours.
Thus, the required number of ways
= 3 ¥ 2 ¥ 2 ¥ 2 = 24.
7!
4! 2!
ways. Corresponding to each arrangement of these seven
letters, we have 8 places where S can be arranged as shown
below with X.
X X X X X X X
We can choose 4 places out of 8 in 8C4 ways.
Thus, the required number of ways
Ê 7! ˆ
8
6
= (8C4) ÁË
˜ = (7) ( C4 )( C4 )
4 ! 2 !¯
Example 74: From 6 different novels and 3 different
dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on a shelf so that the dictionary is always
in the middle. Then the number of such arrangement is
(a) at least 750 but less than 1000
(b) at least 1000
(c) less than 500
(d) at least 500 but less than 750
Ans. (b)
Solution: We can choose 4 novels out of 6 in 6C4 ways
and 1 dictionary out of 3 in 3C1 ways. We can arrange 4
novels and 1 dictionary in the middle in 4! ways. Thus,
required number of ways
= (6C4) (3C1) (4!) = 1080 > 1000
Example 75: There are two urns. Urn A has 3 distinct
red balls and urn B has 9 distinct blue balls. From each urn
two balls are taken out and then transferred to the other. The
number of ways in which this can be done is
(a) 66
(b) 108
(c) 3
(d) 36
Ans. (b)
Solution: The number of ways of transferring the balls
= (3C2) (9C2) = (3) (36) = 108
Example 76: The number of ways in which the diagram in
Figure can be coloured so that each
of the smaller triangle is painted with
one of the three colours yellow, pink
or green and no two adjacent regions
are painted with the same colour, is
(a) 24
(b) 12
(c) 36
(d) 16
Ans. (a)
Example 77: Four married couples are to be seated in
a row having 8 chairs. The number of ways so that spouses
are seated next to each other, is
(a) 72
(b) 186
(c) 384
(d) 516
Ans. (c)
Solution: Let us denote the four married couples of C1,
C2, C3 and C4. We consider each couple as one unit. We can
permute four units in 4! ways. Each couple can be seated in
2! ways. Thus, the required number of ways is
(4!) (2) (2) (2) (2) = 384.
Example 78: 25 lines are drawn in a plane such that no
two of them are parallel and no three of them are concurrent. The number of points in which these lines intersect, is
(a) 300
(b) 315
(c) 325
(d) 450
Ans. (a)
Solution: The number of points in which the lines can
intersect = the number of ways of choosing two lines out
of 25
25
C2 =
25 ¥ 24
= 300 .
2
Example 79: Letter of the word INDIANOIL are
arranged in all possible ways. The number of permutations
in which A, I, O occur only at odd places, is
(a) 720
(b) 360
(c) 240
(d) 120
Ans. (c)
5!
Solution: A, I, I, I, O can occur at odd places in
3!
ways, and the remaining letters N, N, D, L can be arranged
4!
ways.
at the remaining places in
2!
Ê 5! ˆ Ê 4 ! ˆ
Thus, the required number of ways = Á ˜ Á ˜ = 240 ways
Ë 3! ¯ Ë 2 ! ¯
n
Example 80: Let an = Â
r =0
then bn equals
(a) an
n
a
(c)
2 n
n
1
(
n
Cr
2
)
and bn = Â
(b) nan
(d) 0
r =0
(n - r )
2
( n Cr )
,
Permutations and Combinations 6.17
= (1+10) (1+9)(1+7) = 880
Therefore, number of ways of selecting at least one ball =
880 – 1 = 879.
Ans. (c)
Solution: We have
n
bn = Â
n-r
r =0
Thus,
n
bn + bn = Â
r =0
(
n
Cr
n-r
2
(nC )
2
)
n
=
r =0
n
+
Â
r =0
r
fi
bn =
Â
n
n-r
(
n
Cn - r
)
r =0
n
r
2
(nC )
r
2
=Â
=nÂ
r =0
r
(
n
1
(
n
Cr
2
)
Cr
2
)
= n an
n
a .
2 n
Example 81: A committee of at least three members is
to be formed from a group of 6 boys and 6 girls such that
it always has a boy and a girl.The number of ways to form
such a committee is
(b) 212 – 27 – 35
(a) 211 – 26 – 13
(d) 212 – 27 – 13.
(c) 211 – 27 – 35
Ans. (b)
Solution: Required number of ways
= ( 6C 1 + 6C 2 +
+ 6C 6) ( 6C 1 + 6C 2 +
6
6
6
C 6) – ( C 1) ( C 1)
=(26–1) (26–1) – 36 = 212 – 27 – 35
Example 82: There are 10 points in a plane, out of these
6 are collinear. If N is the number of triangles formed by
joining these points, then
(a) N £ 100
(b) 100 < N £ 140
(c) 140 < N £ 190
(d) N > 190.
Ans. (a)
Solution: Number of triangles
=10C3 – 6C3 = 120 – 20 = 100
Example 83: Assuming the balls to be identical except
for difference of colours the number of ways in which one
or more ball can be selected from 10 white, 9 green and 7
black balls is:
(a) 629
(b) 630
(c) 879
(d) 880
Ans. (c)
Solution: We can select 0 or more balls out of 10 white
balls in 11 ways. [select 0 or 1 or 2
or 10 white balls.]
Thus, the number of ways of making a selection of 0 or
more balls
Example 84: Let an = number of n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them is 0.
Let bn = number of such n digit integers ending with digit
1, and cn = the numner of such n digit integers ending with
digit 0, then for n ≥ 3,
(b) an = bn–1 + cn–1
(a) an = bn–1 + cn–2
(c) an = bn–2 + cn–1
(d) an = bn + cn
Ans. (d)
Solution: Note that all such numbers begin with 1.
If number ends in 1, then there bn = an–1 such numbers. If
number ends in 0, then (n–1)th digit cannot be 0 and thus,
there are cn = an–2 such numbers.
\
an = bn + cn = an–1 + an–2
Example 85: In Example 84, a8 equals
(a) 21
(b) 34
(c) 38
(d) 41
Ans. (b)
Solution: If # (S) = number of elements of S, then
a1 = 1, a2 = # {10,11} = 2,
a3 = # {101,110, 111} = 3
Next,
an = bn + cn = an–1 + an–2 " n ≥ 3,
we get
a4 = a3 + a2 = 5,
a5 = a4 + a3 = 8, a6 = 13, a7 = 21
and
a8 = 34
Example 86: The number of natural numbers less than one
million that can be formed by using the digits, 0, 2, and 3 is
(a) 728
(b) 726
(c) 730
(d) 732
Ans. (a)
Solution: The number of k digit natural numbers formed
by using the digit 0,2 and 3 is 2(3k–1).
\ Number of natural numbers less than one million formed
by 0,2 and 3 is
= 2 + (2) (3) + (2) (32) + 2(33) + 2(34) + 2(35)
= 728
Assertion-Reason Type Questions
Example 87:
Statement-1:
The number of ways of distributing 10 identical balls in 4
distinct boxes such that no box is empty is 9C3
Statement-2: The number of ways of choosing 3 places
from 9 different place is 9C3.
Ans. (b)
6.18
Complete Mathematics—JEE Main
Solution: Let xi = number of balls put in ith box,
then x1 + x2 + x3 + x4 = 10 where xi ≥ 1.
Put
xi = yi + 1, so that equation becomes
y1 + y2 + y3 + y4 = 6. where yi ≥ 0
Number of non-negative integral solution of the above equation
= Number of ways of arranging 6 identical balls and 3 identical separators
(6 + 3)! 9
=
= C3 number of ways of choosing 3
6 ! 3!
places out of 9 different places.
n
n
Ê nˆ Ê n – j ˆ
= Á ˜Á
Ë j ¯ Ë k – j ˜¯
n
Thus,
Ê nˆ Ê k ˆ
n
=
Â
\
j =1
j =1
=
n
Ê nˆ Ê k ˆ
 ÁË k ˜¯ ÁË j ˜¯ =
k= j
k= j
Ên –
l
l=0
 ÁË
n
jˆ
n –j
˜¯ = 2
Ê nˆ
 ÁË j ˜¯ 2n – j = (2 + 1) n – 2n
j =1
n
Statement-2 is false as
2
 ( n Ck )
= 2nCn –1
k =1
n
Example 91: Statement-1: Tn =
 k ( 2 n Ck ) = 22n – 2
k =1
n
Statement-2:
n
Solution: 2 Â
\ Statement-2 is true.
2n + 1
Ck =
k=0
For statement-1, put m = n.
Example 89: Let n ΠN, and
n
n +1
Pn
n +1
Cn
n +1
=
n+2
Pn + 2
Cn + 1
n+2
Cn + 2
Statement-1: f (n) is an integer for all n ΠN.
Statement-2: If elements of a determinant are integers, then
determinant itself is an integer.
Ans. (a)
Solution: If each element of a determinant is an integer,
then its each cofactor is an integer, and hence determinant
itself is an integer.
= 22n
n
n
k=0
k=0
 2 n + 1 Ck +  2 n + 1 Ck
n
n
k=0
k=0
 2 n + 1 C k +  2 n + 1 C2 n + 1 – k
2n + 1
n+2
Pn + 1
 2 n + 1 Ck
k=0
Ans. (d)
n
Ê n – jˆ
= 3n – 2n
(mn )!
(n!)m
n
n
Ê nˆ
 ÁË j ˜¯  ÁË k – j˜¯
n– j
k= j
n!
(mn )!
(mn - n )!
n ! ( mn - n )! n! ( mn - 2 n )! n! 0 !
(mn )!
m ! ( n !)m
f (n) =
Ê n – jˆ
 ÁË k – j˜¯
n
As groups are not to be numbered, the desired number of
ways is
n
  ÁË k ˜¯ ÁË j ˜¯
j =1k = j
( mnCn) (mn –nCn ) º (nCn )
=
Cn
n!
(n - j )!
j ! ( n - j )! ( k - j )! ( n - k )!
=
But
=
2n
n!
k!
Ê nˆ Ê k ˆ
Solution: Á ˜ Á ˜ =
Ë k ¯ Ë j ¯ k ! ( n - k )! j ! ( k - j )!
n
Solution: The number of ways of selecting students for
the first group is mnCn ; for the second group is mn –nCn and
so on.
\ the number of ways of dividing (mn) students into m
numbered groups is
=
Ans. (c)
Statement-2: The number of ways of dividing mn students
Ans. (a)
= 3n – 2n
k =1
is a natural number.
(mn )!
into m groups each containing n students is
m
m ! ( n !)
2
 ( n Ck )
Statement-2:
2
(n!)n + 1
Ê nˆ Ê k ˆ
j =1 k = j
Example 88: Statement-1: If n is a natural number then
(n )!
n
  ÁË k ˜¯ ÁË j ˜¯
Example 90: Statement-1:
=
Â
2n + 1
Ck = 22n +1
k=0
n
fi
 2 n + 1 Ck
=2 2n
k=0
\ Statement-2 is true.
Using k ( 2nCk ) = 2n(2n – 1Ck – 1), for k ≥ 1, we get
n
n –1
Tn = 2 n  ( 2 n – 1 C k – 1 ) = 2 n Â
k =1
= 2n(22n –2) = n(22n –1)
j=0
2n – 1
Cj
Permutations and Combinations 6.19
Example 92: Statement-1: The number of ways of distributing at most 12 toys to three children A1, A2 and A3 so that
A1 gets at least one, A2 at least three and A3 at most five, is 145.
Statement-2: The number of non-negative integral solutions of x1 + x2 + x3 £ m is m – 1P2.
Ans. (d)
Solution: Suppose Ai gets xi toys, then
x1 + x2 + x3 £ 12.
Let
x4 = 12 – (x1 + x2 + x3), then
x1 + x2 + x3 + x4 = 12
(1)
The number of non-negative integral solutions of (1)
= coefficient of t 12 in (t + t2 + º) (t 3 + t 4 + t 5 + º)
(1 + t + º + t 5) ¥ (1 + t + t 2 + º)
= coefficient of t 12 in t4(1 – t6) (1 – t) – 4
= coefficient of t 8 in (1 – t6) (1 + 4C1t + 5C2t 2 + º)
= 11C8 – 5C2 = 165 – 10 = 155
Statement-2 is false as the number of non-negative integral
solutions of
x1 + x2 + x3 £ m
equals the number of non-negative integral solution of
x1 + x2 + x3 + x4 = m,
which equals m + 3Cm.
LEVEL 2
Straight Objective Type Questions
Example 93: The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words
that appear before COCHIN is
(a) 360
(b) 192
(c) 96
(d) 48
Ans. (c)
Solution: Letters appearing in the word COCHIN are
C, C, H, I, N, O
Words appearing before COCHIN are of the form
CX---where X is one of the letters C, H, I, N and the four remaining places can be filled by the remaining four letters.
Thus, the number of words before COCHIN is (4) (4!) = 96
Example 94: If r, s, t are prime numbers and p, q are
natural numbers such that LCM of p, q is r 2 s 4 t 2, then the
number of ordered pairs ( p, q) is
(a) 252
(b) 254
(c) 225
(d) 224
Ans. (c)
Solution: Numbers p and q must be of the form
p = r a s b t c, q = r a s b t g
where
0 £ a, a £ 2 and at least one of a, a is 2
0 £ b, b £ 4 and at least one of b, b is 4
0 £ c, g £ 2 and at least one of c, g is 2
Possible values of (a, a), and (c, g) are
(0, 2), (1, 2), (2, 2), (2, 0), (2, 1).
Possible values of (b, b) are
(0, 4), (1, 4), (2, 4), (3, 4), (4, 4), (4, 0), (4, 1), (4, 2), (4, 3).
Thus, number of possible order pairs ( p, q) is 5 ¥ 9 ¥ 5 =
225.
Example 95: If n > 1 and n divides (n – 1)! + 1, then
(a)
(b)
(c)
(d)
n must be prime
n must be divisible by exactly two primes
n must be a composite number
none of these.
Ans. (a)
Solution: If n is not prime, then there exists r ΠN such
that 2 £ r £ n – 1 and r | n.
As r |n and n | [(n – 1)! + 1], we get
r | [(n – 1)! + 1]
As 2 £ r £ n – 1, r | (n – 1)!, therefore r | 1. A contradiction.
Example 96: Which of the following statement is false?
(a) There exist 100 consecutive natural numbers none
of which is prime
(b) There exist 1000 consecutive natural number none
of which is prime.
(c) Given n ΠN, there exist n consecutive natural
numbers none of which is prime.
(d) none of these.
Ans. (d).
Solution: Given any n ΠN, then
(n + 1)! + 2, (n + 1)! + 3, ºº, (n + 1)! + (n + 1)
are n consecutive natural numbers none of which is prime.
6.20
Complete Mathematics—JEE Main
Example 97: The results of 21 football matches (win,
lose or draw) are to be predicted. The number of forecasts
that contain exactly 18 correct results is
(b) 21C18 23
(a) 21C3 218
21
18
(c) 3 – 2
(d) 21C3 321 – 218
Ans. (b)
Solution: 18 correct results can be predicted in
C18
ways and 3 wrong results in 2 ways. Thus, required number
of ways is 21C18 23.
Example 98: There are 15 points in a plane, no two of
which are in a straight line except 4, all of which are in a
straight line. The number of triangles that can be formed by
using these 15 points is
(a) 404
(b) 415
(c) 451
(d) 490
Ans. (c)
Solution: The required number of triangle is
C3 – 4C3 = 455 – 4 = 451
Example 99: If x, y Π(0, 30) such that
5
È x ˘ È 3 x ˘ È y ˘ È 3 y ˘ 11
ÍÎ 3 ˙˚ + ÍÎ 2 ˙˚ + ÍÎ 2 ˙˚ + ÍÎ 4 ˙˚ = 6 x + 4 y
(where [x] denotes greatest integer £ x), then number of
ordered pairs (x, y) is
(a) 0
(b) 2
(c) 4
(d) none of these
Ans. (d)
Solution: Let {x} = x – [x] denote the fractional part of
x. Note that 0 £ {x} < 1. We can write the given equation as
{}
{ } {} { }
{} { } {} { }
11
5
3x
3x
3y
3y
x
x
y
y
= x+ y
+
+ +
6
4
3
3
2
2
2
2
4
4
3x
3y
x
y
+
+
+
=0
3
2
2
4
fi
=
+
As each number on the L.H.S. lies in the interval 0 £ x < 1,
we must have
3x
3y
x
y
=
=
=
=0
3
2
2
4
{} { } {} { }
 Â
10
C j ( j Ci )
0 £ i < j £10
C1 ( C0 + 1C1) + 10C2 (2C0 + 2C1 + 2C2)
… + 10C10 (10C0 + 10C1 + … + 10C10)
10
C1(2) + 10C2(22) + 10C3(23) + … + 10C10(210)
10
10
10
1
=
= (2 + 1) –1 = 3 – 1.
21
3
15
Solution: We have
Example 101: There are p letters a, q letters b, r letters
c. The number of ways of selecting k letters out of these if
p < k < q < r is
1
(p + 1)2– k
2
1
(p + 1) (2k – p)
(b)
2
1
(c)
(p + 1) (q + 1) (r + 1) – k
3
(d) none of these
Ans. (b)
(a)
Solution: Let x a¢s, y b¢s and z c¢s be selected. Then
number of selections is equal to the number of non-negative
integral solutions of
x+y+z=k
(1)
If we take x = l, 0 £ x £ p then y + z = k – l and its number
of solutions is k – l + 1.
Thus, the desired number of selections
p
 (k - l + 1) = (k + 1) (p + 1) –
l =0
1
(p + 1) (p + 2)
2
1
(p + 1) (2k – p).
2
Example 102: The number of ways of choosing n objects out of (3n + 1) objects of which n are identical and
(2n + 1) are distinct, is
(a) 22n
(b) 22n+1
2n
(c) 2 – 1
(d) none of these
Ans. (a)
=
Solution: If we choose k (0 £ k £ n) identical objects,
then we must choose (n – k) distinct objects. This can be
done in 2n+1Cn–k ways.Thus, the required number of ways
n
fi
\
fi
x 3x y
3y
must be integers.
, , and
3 2 2
4
x = 6, 12, 18, 24,
y = 4, 8, 12, 16, 20, 24, 28
Number of ordered pairs (x, y) equals 4 × 7 = 28.
Example 100: The sum
ÂÂ
(10 C j )( j Ci ) is equal to
0 £i £ j £10
10
(a) 2 – 1
(c) 310 –1
Ans. (c)
(b) 210
(d) 310
=
 2 n+1Cn- k
= 2n +1Cn + 2n +1Cn – 1 + … + 2n +1C0 = 22n.
k =0
Example 103: The greatest common divisor
31
…, C29 is
(a) 31
(c) 17
Ans. (a)
(b) 2
(d) none of these
Solution: We have, for 3 £ r £ 29
r! (31 – r)! (31Cr) = 31!
31
C3, 31C5,
Permutations and Combinations 6.21
As the prime 31 divides R.H.S. and 31 does not divide r! and
(31 – r)! for 3 £ r £ 29, we get 31˙ (31Cr).
Also, since 31C29 = (31)(3)(5), 31C3 = (31)(29)(5) and 31C5
= (31)(29)(7)
No prime other than 31 can divide all the numbers.
Thus greatest common divisors of the given numbers is 31.
Example 104: For x Œ R, let [x] denote the great1 ˘
È 1˘ È 1
est integer £ x, then value of Í- ˙ + Í- Î 3 ˚ Î 3 100 ˙˚
2 ˘
È 1
È 1 99 ˘
+ Í- + … + Í- is
˙
Î 3 100 ˚
Î 3 100 ˙˚
(a) – 100
(c) – 135
Ans. (c)
\
1
1
r
1 67
- -1£ - £- 3
3 100
3 100
\
r ˘
È 1
ÍÎ- 3 - 100 ˙˚ = – 2 for 67 £ r £ 100
100
Hence,
n
Also, for 67 £ r £ 100,
for 0 £ r £ 66
r
67
r
67
£
£1 fi–1£ £100 100
100
100
r
= 67(– 1) + 2(– 34) = – 135.
Example 105: If n £ 8 and nCr + 1 = 20 and n – 1Cr = 10,
then n is equal to
(a) 5
(b) 6
(c) 7
(d) 8
Ans. (b)
Solution
r
2
r
2
<
fi- <£0
3
100
100 3
1 2
1
r
1
- - <- £3 3
3 100
3
r ˘
È 1
ÍÎ- 3 - 100 ˙˚ = – 1
1
 ÈÍÎ- 3 - 100 ˘˙˚
r =0
(b) – 123
(d) – 153
Solution: For 0 £ r £ 66, 0 £
fi
fi
Cr +1
n -1
Cr
=
20
10
n!
r !(n - r - 1)!
=2
(r + 1)! (n - r - 1)!
(n - 1)!
fi
n
=2
r +1
This shows that n is even and r + 1 = n/2 Now. nCn/2 = 20
and n £ 8
fi
n = 6.
fi
EXERCISE
Concept-based
Straight Objective Type Questions
(2 x )!
x!
1. If x ΠN and
:
= 44 : 3, then x is
3!(2 x - 3)! 2!( x - 2)
equal to
(a) 6
(b) 7
(c) 11
(d) 12
( x + 2)! (2 x + 1)! 72
= , then x is equal to
2. If x ΠN and
.
7
(2 x - 1)! ( x + 3)!
(a) 3
(b) 4
(c) 5
(d) 6
m+n
m–n
P2 = 90 and
P2 = 30, then (m, n) is equal to
3. If
(a) (12, 8)
(b) (8, 2)
(c) (6, 4)
(d) (7, 4)
4. The number of ways in which you can put five beads
of five different colours to form a necklace is:
(a) 12
(b) 24
(c) 60
(d) 120
5. 15 lines are drawn in a plane in such a way that no
two of them are parallel and no three are concurrent.
The number of points of intersections of these lines is
(a) 455
(b) 465
(c) 475
(d) 485
6. Suppose p ΠN, n = pC2 and m = nC2. If 4m : n =
18 : 1, then p is equal to
(a) 11
(b) 9
(c) 7
(d) 5
7. If 2nC4,
(a) 14
(c) 7
2n
C5 and
2n
C6 are in A.P, then n is equal to
(b) 12
(d) 6
8. If nPr = 1680 and nCr = 70, then n is equal to
(a) 5
(b) 7
(c) 8
(d) 10
9. If n + 1Cr + 1 : nCr :
of n is
(a) 12
(c) 18
n–1
Cr – 1 = 11 : 6 : 3, then value
(b) 10
(d) 21
10. Suppose P is a set containing n distinct elements. Let
S = {(x, y, z.) |x, y, z, Œ, P and at least two of x, y,
z are equal}.
The number of elements in S is
(a) n(3n – 2)
(b) n2 (n – 2)
3 n
(d) n (5n – 4)
(c) n – C3
Complete Mathematics—JEE Main
6.22
LEVEL 1
Straight Objective Type Questions
11. A man invites 6 non-vegeterian and 5 vegeterian friends
for a dinner party. He arrange 6 non-vegeterian friends
on one round table and 5 vegeterian friends along
another round table. The number of ways this can
be done is:
(a) 11!
(b) 9!
(c) 2880
(d) 8280
12. In an examination, there are 11 papers. A candidate
has to pass in at least 6 papers to pass the examination. The number of ways in which the candidate can
pass the examination is:
(b) 210
(a) 29
11
(d) 211 – 11C5
(c) 2
13. At an election, a voter may vote for any number of
candidates not greater than the number to be chosen.
There are 10 candidates and 5 members to be selected.
The number of ways of in which a voter can vote is:
(a) 630
(b) 635
(c) 637
(d) 639
14. In a class of 10 students there are 3 girls students.
The number of ways in which they can be arranged
in a row, so that no two girls are consecutive is k(8!),
then k is equal to
(a) 12
(b) 24
(c) 36
(d) 42
15. The number of ways of putting 5 identical balls in 10
identical boxes, if not more than one can go in a box is
(b) 10C5
(a) 10P5
5
(c) 10
(d) 1
16. The number of four digit numbers that do not contain
4 different digits is:
(a) 4464
(b) 3680
(c) 4120
(d) 7208
17. The number of ways of dividing 10 girls into two
groups of 5 each so that two shortest girls are in the
different groups is:
(a) 70
(b) 252
(c) 140
(d) 282
18. The number of solutions of
1 1 1
+ = where x, y ΠN
x y 6
is
(a) 9
(c) 21
19. Let m and n be two digit natural numbers. The number
of pairs (m, n) such that n can be subtracted from m
without borrowing is:
(a) 2475
(b) 2550
(c) 2675
(d) 2875
20. 5-digit numbers are formed using 2, 3, 5, 7, 9 without repeating the digits. If p is the number of such
numbers that exceeds 20000 and q be the number of
those that lie between 30000 and 90000 then p : q is:
(a) 5 : 4
(b) 5 : 3
(c) 8 : 3
(d) 15 : 16
21. The number of diagonals of a polygon of 15 sides is
(a) 105
(b) 90
(c) 75
(d) 60
22. A five digit number divisible by 3 is to be formed
using the digits 0, 1, 2, 3, 4 and 5 without repetition.
The total number of ways, this can be done is
(a) 240
(b) 3125
(c) 600
(d) 216
23. The number of subsets of a set containing n distinct
object is
(a) nC1 + nC2 + nC3 + … + nCn–1
(b) 2n – 1
(c) nC0 + nC1 + … + nCn
(d) 2n + 1
24. If a, b and r are positive integers, then value of
a
(a)
(c)
Cr + aCr –1 bC1 + aCr– 2bC2 + … + bCr is
a ! b!
r!
a +b
Cr
(d)
(a + b)!
r!
ab
Cr
25. The number of 10 digit numbers that can be written
by using the digits 0 and 1 is
(b) 210
(a) 10C2 + 9C2
10
(c) 2 –2
(d) 10!
26. If Pr stands for r Pr , then sum of the series 1 + P1 +
2P2 + 3P3 + … + nPn is
(a) (n + 1)!
(b) Pn +1 – 1
(d) none of these
(c) Pn – 1 + 1
n
27. If an =
(b) 18
(d) 28
(b)
n
1
 nC
r =0
r
, then value of
Â
r =0
n - 2r
n
Cr
is
Permutations and Combinations 6.23
n
a
2 n
(c) nan
(a)
1
an
4
(d) 0
(b)
28. m men and w women are to be seated in a row so
that all women sit together. The number of ways in
which they can be seated is
(a) (m + 1)! w!
(b) m! w!
(c) m! (w – 1)!
(d) m + wCw
29. A five digit number divisible by 6 is to be formed by
using the digits 0, 1, 2, 3, 4 and 8 without repetition.
The total number of ways in which this can be done
is
(a) 216
(b) 150
(c) 116
(d) 98
30. Rakshit is allowed to select (n + 1) or more books
out of (2n +1) distinct books. If the number of ways
in which he may not select all of them is 255, then
value of n is
(a) 3
(b) 4
(c) 5
(d) 11
31. The number of ways in which a committee of 3
women and 4 men be chosen from 8 women and 7
men if Mr. X refuses to serve on the committee if Mr.
Y is a member of the committee is
(a) 420
(b) 840
(c) 1540
(d) 1400
32. The product of first n odd natural numbers equals
(a) (2nCn) (nPn)
Ê 1ˆ n
(b) Á ˜ (2nCn) (nPn)
Ë 2¯
n
Ê 1ˆ
(c) Á ˜ (2nCn) (2nPn)
Ë 4¯
(d) none of these
33. The number of ways in which two teams A and B of
11 players each can be made up from 22 players so
that two particular players are on the opposite sides is
(a) 369512
(b) 184755
(c) 184756
(d) 369514
34. The number of ways in which 20 letters x1, x2, … x10,
y1, y2, …, y10 be arranged in a line so that suffixes
of the letters x and also those of y are respectively in
ascending order of magnitude is
(a) 126
(b) 64
20
(d) 184756
(c) 2
35. The product of n consecutive natural number is always
divisible by
(b) 2nCn
(a) nPn
2n
(c) Pn
(d) n +1Pn
36. Six X have to be placed in the squares of Figure such
that each row contain at least one X. The number of
ways in which this can be done is
(a) 25
(b) 26
(c) 27
(d) 30
37. There are three pigeon holes marked M, P, C. The
number of ways in which we can put 12 letters so
that 6 of them are in M, 4 are in P and 2 are in C is
(a) 2520
(b) 13860
(c) 12530
(d) 25220
38. The greatest number of points of intersection of n
circles and m straight lines is
(a) 2mn + mC2
1
m(m – 1) + n(2m + n – 1)
2
(b)
(c) mC2 + 2(nC2)
(d) none of these
39. The number of binary
contain even number of
(a) 2n–1
(c) 2n – 2
sequences of length n that
1’s is
(b) 2n – 1
(d) none of these
40. The number of natural numbers with distinct digits is
(b) 1010 – 910
(a) 910–1
10
(c) 9 ¥ 1
(d) none of these
41. The number of five digit numbers that contain 7 exactly once is
(b) (37) (93)
(a) (41) (93)
4
(d) (41) (94)
(c) (7) (9 )
42. The units digit of 172009 + 112009 – 72009 is
(a) 1
(b) 8
(c) 2
(d) 9
43. If
1
4
Cn
(a) 3
(c) 1
=
1
5
Cn
+
1
6
, then value of n is
Cn
(b) 4
(d) 2
44. At an election there are five candidates and three
members are to be elected, and a voter may vote for
any number of candidates not greater than the number
Complete Mathematics—JEE Main
6.24
to be elected. The number of ways in which the person
can vote is
(a) 25
(b) 30
(c) 35
(d) 25–23
45. If n is odd and nC0 < nC1 < nC2 < … < nCr, then
maximum possible value of r is
n -1
n +1
(b)
(a)
2
2
(c) n
(d) none of these
52. The number of five digit numbers that can be formed
by using digits 1, 2, 3 only, such that three digits of
the formed number are identical, is
(a) 30
(b) 60
(c) 120
(d) 90
n
ÈS ˘
Sn - 7 Í n ˙ , where [x] denotes
Î7˚
r =1
greatest integer £ x, then T100 is equal to
(a) 7
(b) 5
(c) 11
(d) none of these
53. Let Sn =
 r ! . If Tn =
46. The number of arrangements of the letters of the word
BANANA in which two N’s do not appear adjacently
is
(a) 40
(b) 60
(c) 80
(d) 100
54. The number of ordered pairs (m, n), m, n Π{1, 2,
…, 50} such that 6m + 9n is a multiple of 5 is
(a) 1250
(b) 2500
(c) 500
(d) 625
47. The number of (staircase) paths in the xy-plane from
(0, 0) to (7, 5) where each such path is made up of
individual steps going one unit upward (U) or one
unit to the right (R). One such path is shown in figure
(b) 12!
(a) 12C5
(c) 5! 7!
(d) 12P5
55. Ten different letters of an alphabet are given. Words
with five letters are formed from these given letters.
The number of words which have at least one of their
letter repeated is
(a) 69760
(b) 30240
(c) 99748
(d) none of these
56. The number of squares that we can find on a chess
board is
(a) 64
(b) 160
(c) 224
(d) 204
y
5
4
57. In a football championship, 153 matches were played.
Every team played one match with each other, the
number of teams participating in the championship is
(a) 18
(b) 16
(c) 9
(d) none of these
3
2
1
49. The number of ways we can put 5 different balls in 5
different boxes such that at most three boxes is empty,
is equal to
(b) 55–10
(a) 55– 5
(d) none of these
(c) 25–2
58. The total number of ways of dividing mn distinct
objects into n equal groups is
(mn)!
(mn)!
(b)
(a)
n
(m !) (n!)
(n!)m (m !)
(mn)!
(c)
(d) none of these
m ! n!
59. The number of ways in which we can get at least 6
successive tails when a coin is tossed 10 times in a
row, is
(b) 27 – 26
(a) 10C6
(c) 40
(d) 48
50. The number of ordered pairs of integers (x, y) satisfying the equation x2 + 6x + y2 = 4 is
(a) 2
(b) 4
(c) 6
(d) 8
60. Let R = {a, b, c, d} and S = {1, 2, 3}, then the number
of functions f, from R to S, which are onto is
(a) 80
(b) 16
(c) 24
(d) 36
51. The number of the factors of 20! is
(a) 4140
(b) 41040
(c) 4204
(d) 81650
61. If n +1C2 + 2(2C2 + 3C2 + 4C2 + … + nC2 ) = x, then
x equals
(a) 1 + 2 + 3 … + n
(b) 12 + 22 + … + n2
1
(c)
n(n + 1) (n + 3) (d) none of these
2
O
1
2
3
4
5
6
7
x
48. In Question 47, how many such paths are there if
each path must pass through the point (3, 4)?
(a) 175
(b) 145
(c) 95
(d) 78
Permutations and Combinations 6.25
62. The number of ways in which three girls and nine
boys can be seated in two cars, each having numbered
seats, 3 in the front and 4 at the back, is
(b) 14P12 × 14 × 3
(a) 14P12
(c)
11
P9
(d) none of these
63. The number of integral solutions of x + y + z = 0
with x ≥ – 5, y ≥ – 5, z ≥ – 5 is
(a) 134
(b) 136
(c) 138
(d) 140
64. A library has n different books and has p copies of
each of the book. The number of ways of selecting
one or more books from the library is
(b) pn
(a) np
p
(c) (n + 1) – 1
(d) (p + 1)n – 1
65. The number of ways in which we can get a score of
11 by throwing three dice is
(a) 18
(b) 27
(c) 45
(d) 56
66. Three straight lines l1, l2 and l3 are parallel and lie in
the same plane. Five points are taken on each of l1,
l2 and l3. The maximum number of triangles which
can be obtained with vertices at these points, is
(a) 425
(b) 405
(c) 415
(d) 505
67. The number of 9 digit numbers which have all distinct
digits, is
(a) 10!
(b) 9!
(c) 8(9!)
(d) 10! – 9!
68. A four digit number of distinct digits is formed by
using the digits 2, 3, 4, 5, 6, 7, 8. The number of
such numbers which are divisible by 25, is
(a) 60
(b) 40
(d) 15
(c) 20
69. The range of the function f(x) = 15 – x Px – 8 is
(a) {1, 6, 20, 24}
(b) {1, 6, 10, 15}
(c) {1, 4, 11, 15}
(d) {1, 8, 11, 19}
70. The number of ways of arranging 20 boys so that 3
particular boys are separated is
(a) 9(16!)
(b) 15(16!)
(c) 15(17!)/2
(d) none of these
71. A library has n different books and 3 copies of each of
the n books. The number of ways of selecting one or
more books from the library is
(a) 4n – 1
(b) 5n – 1
n
(d) n3 – 1.
(c) 2 – 1
72. Let an = 10n/n! for n ≥ 1. Then an take the greatest
value when n equals
(a) 20
(b) 18
(c) 6
(d) 9
Assertion-Reason Type Questions
73. Statement-1: The expression
Ê 40ˆ Ê 60ˆ Ê 40 ˆ Ê 60ˆ
ÁË r ˜¯ ÁË 0 ˜¯ + ÁË r – 1˜¯ ÁË 1 ˜¯ + attains maximum value
when r = 50.
Ê 2nˆ
Statement-2: Á ˜ is maximum when r = n.
Ë r¯
74. Statement-1: The number of non-negative integral
solution of
x1 + x2 + º + x20 = 100
120
Ê
ˆ
is Á
.
Ë 20 ˜¯
Statement-2: The number of ways of distributing n
identical objects among r persons giving zero or more
Ê n + r – 1ˆ
.
objects to a person is Á
Ë r – 1 ˜¯
75. Statement-1: The sum of divisors of
n = 210 32 53 72 112 is
1
(211 – 1) (33 – 1) (54 – 1) (73 – 1) (113 – 1)
480
Statement-2: The number of divisors of m =
a
a
p1 1 p2 2 pra r where p1, p2, º pr are distinct primes
and a1, a2, º ar are natural numbers is (a1 + 1) (a2
+ 1) º (ar + 1).
Ê 1000ˆ
is not divisible by 11.
76. Statement-1: The number Á
Ë 500 ˜¯
Statement-2: If p is a prime, the exponent of p in n! is
Èn˘ È n ˘ È n ˘
Í p˙ + Í 2 ˙ + Í 3 ˙ +
Î ˚ Îp ˚ Îp ˚
where [x] denotes the greatest integer £ x.
77. Statement-1: A student is allowed to select at most
n books from a collection of (2n + 1) books. If the
total number of ways in which he can select at least
one book is 255, then n = 3.
2n + 1ˆ Ê 2n + 1ˆ
Statement-2: ÊÁ
+
Ë 0 ˜¯ ÁË 1 ˜¯
Ê 2n + 1ˆ
= 4n
++ Á
Ë n ˜¯
6.26
Complete Mathematics—JEE Main
LEVEL 2
Straight Objective Type Questions
78. 2m white counters and 2n red counters are
arranged in a straight line with (m + n) counters on
each side of a central mark. The number of ways of
arranging the counters, so that the arrangements are
symmetrical with respect to the central mark, is
(a)
(m + n)!
m ! n!
(b)
(c)
1 (m + n)!
2 m ! n!
(d) none of these
2m + 2n
C2m
79. Let x = (2n + 1) (2n + 3) (2n + 5) …
(4n – 3) (4n – 1) and
n
Ê 1 ˆ (4n)! n!
y= Á ˜
, then x – y + 2n is equal to
Ë 2 ¯ (2n)! (2n)!
(a) 0
(c) 2nCn
(b) (2n)!/2n
(d) 2n
80. There are 3 set of parallel lines containing
respectively p lines, q lines and r lines respectively.
The greatest number of parallelograms that can be
formed by the system
(a) pqr + (p – 1) (q – 1) (r – 1)
1
{pqr + (p – 1) (q – 1) (r – 1)}
4
1
(c)
pqr (p + 1) (q + 1) (r + 1)
4
(d) none of these
(b)
81. The number of six digit numbers which have sum of
their digits as an odd integer, is
(a) 45000
(b) 450000
(c) 97000
(d) 970000
82. In the identity
m
A
m!
= Â i , then
x( x + 1)( x + 2)… ( x + m) i = 0 x + i
value of Ak is
(a) nCk
(c) (– 1)k . nCk
(b) nCk +1
(d) none of these
83. The number of integral solutions of x 2 – y2 = 352706,
is
(a) 276
(b) 0
(c) 720
(d) infinite
84. Let A be a set containing n elements. The number of
ways of choosing two subsets P and Q of A such that
P « Q = f is
(b) 3 n
(a) 2 n
n
n
(d) 4 – 2
(d) 4n – 3n.
85. The last digit of (1! + 2! + ºº + 2009!) 500 is
(a) 1
(b) 2
(c) 7
(d) 9
86. If n = 210032 and {d1, d2, º dk} is the set of all divik
1
equals
sors of n, then Â
j =1 dj
(a) 2
(c) 2N
(b) N
(d) none of these
87. If nCr – 1 = (k 2 – 8) ( n + 1 Cr), then k belongs to
(a) [– 3, –2 2 )
(b) (– 3, 3)
(c) [– 3, –2 2 ) » ( 2 2 , 3]
(d) [ –2 2 , 2 2 ]
Previous Years' AIEEE/JEE Main Questions
1. The number of ways in which 6 men and 5 women
can dine at a round table if no two women are to sit
together is given by
(a) 30
(b) 5! ´ 4!
(c) 7! ¥ 5!
(d) 6! ¥ 5!
[2003]
2. A student is to answer 10 questions out of 13 questions in an examination such that he must choose at
least 4 from the first five questions. The number of
choices available to him is
(a) 196
(b) 280
(c) 346
(d) 140
[2003]
Permutations and Combinations 6.27
3. If nCr denotes the number of combinations of n things
taken r at a time, then the expression
n
C r + 1 + nC r – 1 + 2 ¥ nC r
equals
(a) n + 2Cr + 1
(b) n + 1Cr
n+1
(c)
Cr + 1
(d) n + 2 Cr
[2003]
7–x
4. The range of the function
P x – 3 is
(a) {1, 2, 3, 4}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3}
(d) {1, 2, 3, 4, 5} [2004]
5. How many ways are there to arrange the letters in the
word GARDEN with the vowels in alphabetical order?
(a) 360
(b) 240
(c) 120
(d) 480
[2004]
6. The number of ways distributing 8 identical balls in
3 distinct boxes so that none of the boxes is empty
is
(b) 21
(a) 38
[2004]
(c) 5
(d) 8C3
n
7. If Sn =
n
1
 nC
r=0
r=0
r
to
(a) n – 1
(c)
r
 nC
and t n =
1
n
2
, then
r
(b)
1
n–1
2
(d)
2n – 1
2
tn
is equal
Sn
[2004]
6
50
8. The value of
(a)
(c)
56
C3
55
C4
C4 +
 56 – r C3
r =1
(b)
(d)
56
55
C4
C3
is
[2005]
9. If the letters of the word SACHIN are arranged in
all possible ways and these are written out as in a
dictionary, then the word SACHIN appears at serial
number
(a) 603
(b) 602
(c) 601
(d) 600
[2005]
10. At an election, a voter may vote for any number of
candidates, not greater than the number to be elected.
There are 10 candidates and 4 are to be elected. If a
voter votes for at least one candidate, then the number
of ways in which he can vote is
(a) 1110
(b) 5040
(c) 62010
(d) 385
[2006]
11. The set S = {1, 2, 3 º 12} is to be partitioned into
three sets A, B, C of equal size such that, A » B »
C = S, A « B = B « C = C « A = f. The number
of ways to partition S is.
(a)
(c)
12!
3! ( 4!)3
12!
( 4!)3
(b)
(d)
12!
3! (3!)3
12!
(3!)4
[2007]
12. In a shop there are five types of ice-creams available.
A child buys six ice-creams
Statement-1: The number of different ways the
child can buy the six ice-creams is 10C5.
Statement-2: The number of different ways the
child can buy the six ice-creams is equal to the
number of different ways of arranging 6 A’s and
4B’s in a row.
[2008]
13. From 6 different novels and 3 different dictionaries, 4
novels and 1 dictionary are to be selected and arranged
in a row on a shelf so that the dictionary is always in
the middle. Then the number of such arrangements is
(a) at least 750 but less than 1000
(b) at least 1000
(c) less than 500
(d) at least 500 but less than 750.
[2009]
14. There are two urns. Urn A has 3 distinct red balls and
urn B has 9 distinct blue balls. From each urn two
balls are taken out at random and then transferred to
the other. The number of ways in which this can be
done is
(a) 66
(b) 108
(c) 3
(d) 36
[2010]
15. There are 10 points in a place, out of these 6 are
collinear. If N is the number of triangles formed by
joining these points, then
(a) N £ 100
(b) 100 < N £ 140
(c) 140 < N £ 190
(d) N > 190
[2011]
16. Statement-1: The number of ways of distributing 10
identical balls in 4 distinct boxes such that no box is
empty is 9C3.
Statement-2: The number of choosing 3 places from
[2011]
9 different places is 9C3.
17. Assuming the balls to be identical except for difference in colours, the number of ways in which one or
more balls can be selected from 10 white, 9 green
and 7 black balls is:
(a) 629
(b) 630
(c) 879
(d) 880
[2012]
18. Let Tn be the number of all possible triangles formed
by joining vertices of an n-sided regular polygon. If
Tn + 1 – Tn = 10, then value of n is
(a) 5
(b) 10
(c) 8
(d) 7
[2013]
6.28
Complete Mathematics—JEE Main
19. The number of ways in which an examiner can assign
30 marks to 8 question, giving no less then 2 marks
to any question is:
(b) 21C8
(a) 30C7
21
(c) C7
(d) 30C8
[2013, online]
27. Let A and B be two sets containing four and two
elements respectively. Then the number of subsets of
the set A ¥ B, each having at least three elements is:
(a) 219
(b) 256
(c) 275
(d) 510
[2015]
20. 5-digit numbers are formed using 2, 3, 5, 7, 9 without repeating the digits. If p is the number of such
numbers that exceeds 20000 and q be the number of
those that lie between 30000 and 90000, then p : q
is:
(a) 6 : 5
(b) 3 : 2
(c) 4 : 3
(d) 5 : 3
[2013, online]
28. The number of integers greater than 6,000 that can
be formed, using the digits 3, 5, 6, 7 and 8, without
repetition, is:
(a) 216
(b) 192
(c) 120
(d) 72
[2015]
21. A committee of 4 persons is to be formed from 2
ladies, 2 old men and 4 youngmen such that it includes
at least 1 lady, at least one old man and at most 2
youngmen. Then the total number of ways in which
this committee can be formed is
(a) 40
(b) 41
(c) 16
(d) 32
[2013, online]
22. On the sides AB, BC, CA of a triangle ABC, 3, 4, 5
distinct points (excluding vertices A, B, C) are respectively chosen. The number of triangles that can be
constructed using these chosen points as vertices are:
(a) 210
(b) 205
(c) 215
(d) 220
[2013, online]
23. The sum of the digits in the unit’s place of all the
4-digit numbers formed by using the numbers 3, 4,
5 and 6, without repetition, is:
(a) 432
(b) 108
(c) 36
(d) 18
[2014, online]
24. An eight digit number divisible by 9 is to be formed
using digits from 0 to 9 without repeating the digits.
The number of ways in which this can be done is:
(a) 72(7!)
(b) 18(7!)
(c) 40(7!)
(d) 36(7!) [2014, online]
25. 8-digit numbers are formed using the digits 1, 1, 2,
2, 2, 3, 4, 4. The number of such numbers in which
the odd digits do not occupy odd places, is:
(a) 160
(b) 120
(c) 60
(d) 48
[2014, online]
26. Two women and some men participated in a chess
tournament in which every participant played two
games with each of the other participants. If the number of games that the men played betweeen themselves
exceeds the number of games that the men played
with the women by 66, then the number of men who
participated in the tournament lies in the interval:
(a) [8, 9]
(b) [10, 12)
(c) (11, 13]
(d) (14, 1) [2014, online]
* None of the given options is correct
29. The number of points, having both co-ordinates as
integers, that lie in the interior of the triangle with
vertices (0, 0), (0, 41) and (41, 0), is:
(a) 901
(b) 861
(c) 820
(d) 780
[2015]
30*. The number of ways of selecting 15 teams from 15
men and 15 women, such that team consists of a man
and a woman, is:
(a) 1120
(b) 1240
(c) 1880
(d) 1960
[2015 online]
31. If in a regular polygon the number of diagonals is 54,
then the number of sides of this polygon is:
(a) 10
(b) 12
(c) 9
(d) 16
[2015 online]
32. If all the words (with or without meaning) having five
letters, formed using the letters of the word SMALL
and arranged as in a dictionary, then the position of
the word SMALL is:
(b) 59th
(a) 46th
nh
(c) 52
(d) 58th
[2016]
15
Ê 15 C ˆ
33. The value of  r 2 Á 15 r ˜ is equal to:
Ë Cr -1 ¯
r =1
(a) 1240
(c) 1085
(b) 560
(d) 680
[2016 online]
15
34. The sum  (r 2 + 1) ¥ (r !) is equal to:
r =1
(a) 11 ¥ (11!)
(c) (11)!
(b) 10 ¥ (11!)
(d) 101 ¥ (10!)
[2016, online]
35. If the four letter words (need not be meaningful) are
to be formed using the letters from the word “MEDITERRANEAN” such that the first letter is A and the
fourth letter is E, then the total number of all such
words is:
(a) 110
(b) 59
11!
(d) 56
[2016, online]
(c)
(2!)3
Permutations and Combinations 6.29
36.
If
2
n+2
C6
n-2
P2
= 11, then n satisfies the equation:
(b) n2 + 2n – 80 = 0
(d) 4n2 + 5n – 84 = 0
[2016, online]
(a) n + n – 110 = 0
(c) n2 + 3n – 108 = 0
Previous Years' B-Architecture Entrance
Examination Questions
1. A set B contain 2007 elements. Let C be the set consisting of subsets of B which contain at most 1003
elements. The number of elements in C is:
(b) 22006
(a) 22005
1003
(c) 2
(d) 22007
[2006]
2. A parallelogram is cut by two set of n parallel lines,
parallel to the sides of the parallelogram. The number
of parallelograms formed is:
(b) (nC2) (n + 1C2)
(a) (n + 2C2) (n + 2C2)
n
n+2
(c) ( C2) ( C2)
(d) (nC2) (nC2)
[2007]
3. If there are 30 onto, mappings from a set containing
n elements to the set {0, 1}, then n equals:
(a) 3
(b) 5
(c) 7
(d) 2
[2007]
4. Let A be a set containing ten elements. Then the number of subsets of A containing at least four elements
is
(a) 845
(b) 848
(c) 850
(d) 854
[2008]
5. A group of 2n students consisting of n boys and n
girls are to be arranged in a row such that adjacent
members are of opposite sex. The number of ways in
which this can be done is:
(a) 2(n!)
(b) (n!)2
2
(c) 2(n!)
(d) n!
[2009]
6. The maximum possible number of points of intersection of 8 straight lines and 4 circles is
(a) 164
(b) 76
(c) 104
(d) 32
[2010]
7. A committee consisting of at least three members is
to be formed from a group of 6 boys and 6 girls such
that it always has a boy and a girl. The number of
ways to form such committee is:
(b) 211 – 26 – 13
(a) 212 – 27 – 13
12
7
(c) 2 – 2 – 35
(d) 211 – 27 – 35
[2011]
8. The total number of injective mappings from a set
with m elements to a set with n elements for, m > n,
is:
m!
n!(m - n)!
(c) nm
m!
(m - n)!
(d) zero
(b)
(a)
[2012]
10
9. The least positive integral value of x for which Cx – 1
> 2(10Cx) is:
(a) 5
(b) 8
(c) 9
(d) 6
[2013]
10. Suppose that six students, including Madhu and Puja,
are having six beds arranged in a row. Further, suppose that Madhu does not want a bed adjacent to Puja.
Then the number of ways, the beds can be allotted to
students is:
(a) 264
(b) 480
(c) 600
(d) 384
[2014]
11. In a car with seating capacity of exactly five persons,
two persons can occupy the front seat and three
persons can occupy the back seat. If amongst the seven
persons, who wish to travel by this car, only two of
them know driving then number of ways in which the
car can be fully occupied and driven by them, is:
(a) 360
(b) 60
(c) 240
(d) 720
[2015]
3
Ê n Ci -1 ˆ
36
12. If S = Â Á n
˜ = 13 , then n is equal to:
n
i =1 Ë Ci + Ci -1 ¯
(a) 11
(b) 12
(c) 13
(d) 10
[2016]
n
13. A code word of length 4 consists two distinct consonants in the English alphabet followed by two digits
from 1 to 9, with repetition allowed in digits. If the
number of code words so formed ending with an even
digit is 432 k, then k is equal to:
(a) 5
(b) 49
(c) 35
(d) 7
[2016]
Answers
Concept-based
1. (a)
5. (a)
9. (b)
2. (b)
6. (d)
10. (a)
3. (b)
7. (c)
4. (a)
8. (c)
6.30
Complete Mathematics—JEE Main
Level 1
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
63.
67.
71.
75.
(c)
(d)
(a)
(c)
(d)
(d)
(a)
(a)
(d)
(a)
(b)
(a)
(d)
(b)
(d)
(a)
(b)
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
64.
68.
72.
76.
(b)
(a)
(b)
(c)
(a)
(b)
(b)
(d)
(a)
(a)
(c)
(d)
(d)
(d)
(b)
(d)
(a)
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
57.
61.
65.
69.
73.
77.
(c)
(a)
(b)
(b)
(b)
(a)
(b)
(a)
(a)
(a)
(b)
(a)
(b)
(b)
(a)
(a)
(d)
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
62.
66.
70.
74.
(d)
(a)
(d)
(a)
(b)
(d)
(b)
(a)
(a)
(d)
(a)
(a)
(a)
(a)
(d)
(d)
Level 2
78. (a)
82. (c)
86. (d)
79. (d)
83. (b)
87. (c)
80. (d)
84. (b)
81. (b)
85. (a)
Previous Years' AIEEE/JEE Main Questions
1.
5.
9.
13.
17.
21.
25.
29.
33.
(d)
(a)
(c)
(b)
(c)
(b)
(b)
(d)
(d)
2.
6.
10.
14.
18.
22.
26.
30.
34.
(a)
(b)
(d)
(b)
(a)
(b)
(b)
(**)
(b)
3.
7.
11.
15.
19.
23.
27.
31.
35.
(a)
(c)
(a)
(a)
(c)
(b)
(a)
(b)
(b)
4.
8.
12.
16.
20.
24.
28.
32.
36.
(b)
(c)
(b)
(c)
2. (a)
6. (c)
10. (b)
3. (b)
7. (a)
11. (d)
Hints and Solutions
Concept-based
1.
(2 x )!
2!( x - 2)! 44
.
=
3!(2 x - 3)!
x!
3
(2 x + 1)(2 x ) 72
=
x+3
7
2
fi 7(2x + x) = 36(x + 3)
fi 14x2 – 29x – 108 = 0
fi (14x + 27) (x – 4) = 0 fi x = 4
3. (m + n) (m + n – 1) = 90 and
(m – n) (m – n – 1) = 30
fi m + n = 10, m – n = 6
\ m = 8, n = 2
2.
4. Number of ways of putting beads in a necklace is 4!
But clockwise arrangement abcde and anticlockwise
arrangement aedcb are not really different, for, the
arrangement abcde gives rise to arrangement aedcb
if the necklace is turned over.
1
\ required number of ways = (4!) = 12
4
5. For a point of intersection, we require two lines,
therefor, the number of points of intersection is 15C3.
1
4m
6. m = n(n – 1) fi
= 2(n – 1) = p( p – 1) – 2
2
n
\ 18 = p( p – 1) – 2 fi p = 5
7. 2(2nC5) =
(c)
(b)
(d)
(b)
(d)
(d)
(b)
(d)
(c)
Previous Years' B-Architecture Entrance
Examination Questions
1.
5.
9.
13.
(2 x )(2 x - 1)(2 x - 2) 44
=
3 x( x - 1)
3
fi 2x – 1 = 11 fi x = 6
fi
4. (b)
8. (d)
12. (b)
2n
fi 2=
2n
2n
C4 +
C4
C5
+
2n
C6
2n
C6
2n
C5
=
5
2n - 5
+
6
2n - 4
fi 12(2n – 4) = 30 + (2n – 4) (2n – 5)
fi 2n2 – 21n + 49 = 0 fi n = 7
8. nPr = (r!) (nCr) fi 1680 = (r!) (70)
fi r! = 24 fi r = 4
Also, 1680 = nP4 = n(n – 1) (n – 2) (n – 3)
fi (n2 – 3n) (n2 – 3n + 2) = 1680
fi (n2 – 3n + 1)2 = 412 fi n2 – 3n + 1 = 41
fi n2 – 3n – 40 = 0 fi (n + 5) (n – 8) = 0
fi n = 8.
9. As (r + 1) (n + 1Cr
we get,
+ 1)
= (n + 1) (nCr),
n + 1 11
= .
r +1 6
n 6
1
= fir = n
r 3
2
n +1
11
= fi n = 10
Thus,
n/2 +1 6
Also,
10. We can choose x, y, z in n3 ways out of which nP3
have distinct elements.
Permutations and Combinations 6.31
Level 1
A number will lie between 30000 and 90000 if it
begins with 3, 5 or 7
11. Arrange non-vegeterians in (6 – 1)! = 5! ways and
vegeterian in (5 – 1)! = 4!ways
21. Number of diagonals =
\ required number of ways
12. Number of ways passing the examination is
11
C6 +
11
C7 +
11
C8 +
11
C9 +
11
C10 +
C2 +
11
C1 +
11
C11
11
C5 +
11
C4 +
11
C3 +
11
When 0 is not used, the number of such numbers is 5!
and when 3 is not used, the number of such numbers
is 5! – 4!
11
C0
(2)
\ Required number of ways = 5! + (5! – 4!) = 216
Adding (1) and (2) we get
2S = 2
11
fiS=2
23. Use: The number of subsets containing exactly r elements is n Cr .
10
13. The voter has to vote for at least one and at most 5
candidates. The number of ways is:
10
C1 +
10
C2 +
10
C3 +
10
C4 +
C2 – 15 = 90.
(1)
Using nCr = nCn – r, we get
S=
15
22. As 0 + 1 + 2 + 3 + 4 + 5 = 15, for a five digit number
to be divisible by 3 either do not use 0 or 3.
= (5!) (4!) = (120) (24) = 2880
S=
\ q = (3) (4!). Thus p : q = 5 : 3.
10
C5 = 637
14. Seven boys can be arranged in 7! ways. If
XB1 XB2 XB3 XB4 XB5 XB6 XB7 X is one such arrangement, then 3 girls can be arranged at any of
the 3 places marked with a X. Thus, the number of
arrangements is (8P3) (7!) = (6 ¥ 7) (8!) = (42) (8!).
Thus, k = 42.
24. The given expression = number of ways of selecting r persons out of a men and b women
= a + bC r
25. For each place we have two choices: either use 0 or
use 1.
26. 1 + P1 + 2P2 + 3P3 + º + nPn
= 1 + 1! + 2(2!) + 3(3!) + º + n(n!)
n
= 1+
r =1
15. There is only one way in which five identical balls
can be put in 10 boxes.
16. The number of 4 digit numbers is (9) (10) (10) (10),
out of these (9) (9) (8) (7) have distinct digits.
17. Excluding two shortest girls, the remaining eight girls
1
can be divided into two groups in (8 C4 )( 4 C4 ) = 35
2!
ways. The two shortest girls can be put in two different groups in 2 ways. Therefore, the required number
of ways is 70.
1 1 1
18. + = fi 6 ( x + y) = xy
x y 6
fi (x – 6) (y – 6) = 36
fi (x – 6)|36.
fi x – 6 = 1, 2, 3, 4, 6, 9, 12, 18, 36
fi (x, y) = (7, 42), (8, 24), (9, 18), (10, 15), (12, 12),
(15, 10), (18, 9), (24, 8), (42, 7)
19. Suppose m = 10a + b and n = 10c + d where 1 £ a, c
£ 9 and 0 £ b, d £ 9. Note that 0 £ d £ b. Thus d can
take (b + 1) values. Similarly , 1 £ c £ a, therefore, c
can take a values. Hence, the required number of pairs
(m, n) is (1 + 2 + . . . + 10) (1 + 2 + .. + 9) = 2475
20. Any number 5-digit formed by 2, 3, 5, 7, 9, exceeds
20000. Therefore, p = 5! = 120
n
 [(r + 1) – 1]r ! = 1 +  {(r + 1)! - r !}
r =1
= 1 + (n + 1)! – 1! = (n + 1)!
n
Â
27.
n – 2r
r=0
n
=
Â
r=0
n
Cr
n–r
n
Cn – r
n
=
Â
n–r
r=0
n
n
–
Â
r=0
Cr
r
n
Cr
n
–
Â
r=0
r
n
Cr
=0
28. Treating w women as one block, we can permute m
men and one block in (m + 1)! ways and women in
the block in w! ways. Thus, the required number of
ways is (m + 1)! w!
29. Since 0 + 1 + 2 + 3 + 4 + 8 = 18, the 5 digit number
will be divisible by 3 if either 0 or 3 is not used.
When 0 is not used. For the unit place we have 3
choices (2, 4 or 8) and for the remaining place we
have 4! Choices.
When 3 is not used
In this case, 0 is used at the unit’s place, the number
of choices is 4! If 0 is not used at the unit’s place,
then unit’s place can be filled up in 3 ways and the
remaining places in (4! – 3!) ways.
Thus, required number of numbers
= 3(4!) + 4! + 3(4! – 3!) = 150
Complete Mathematics—JEE Main
6.32
30. Use
2n + 1
Cn + 1 +
2n + 1
Cn + 2 + º +
2n + 1
C 2n = 22n – 1
31. Women can be chosen in 8C3 ways and men in 7C4 – 5C2
(2 n )!
1
32. 1.3.5 º (2n – 1) = n
=Ê ˆ
2 ( n!) Ë 2 ¯
n
(
2n
Cn )( Pn )
n
33. We can choose 10 players from (22 – 2) players in
20
C10 ways and one player from 2 players in 2C1 ways.
\
20!
¥ 2 = 369512
10! 10!
35. Let n consecutive natural numbers be m + 1, m + 2,
º, m + n where m ≥ 0.
P = (m + 1) (m + 2) º (m + n)
m ! (m + 1) (m + n ) ˆ
= ÊÁ
˜¯ n!
Ë
m ! n!
= ( m + nC n) ( nP n)
P is divisible by nPn
36. This can be done in four mutually exclusive ways as
follows:
Row
R1
Row R2 Row R3 Number of ways
2
4
1
3
2
( C 1) ( C 3 ) ( C 2) = 8
II
1
4
1
( 2C 1) ( 4C 4) ( 2C 1) = 4
III
2
2
2
( 2C 2) ( 4C 2) ( 2C 2) = 6
IV
2
3
1
( 2C 2) ( 4C 3) ( 2C 1) = 8
Total
12
= 1+
Cn
fi
5
Cn
6
Cn
5
6–n
=1+
5–n
6
2m + 1
Cr is maximum when r = m or r = m + 1.
46. Required number of ways
= Total number of ways – Number of ways when two
N’s are together
6!
5!
= 40
=
3! 2! 3!
47. The number of paths
= the number of ways of arranging 5 U ¢ S
and 7 R¢s.
12!
=
= 12C5
5! 7!
48. Required number of paths
= (Number of paths upto (3, 4)) ¥ (Number
of paths from (3, 4) to (5, 7))
=
2
I
37. Number of ways is
C.
4
Cn
45. Use
34. Choose 10 places (out of 20) for x1, x2, º, x10 in
20
C10 ways.
\
5
43.
fi n 2 – 17n + 30 = 0 fi n = 2.
44. Number of ways = 5C1 + 5C2 + 5C3 = 25
required number of ways
=
42. Use unit’s digit of 17 2009 is same as that of the unit’s
digit of 7 2009.
26
C6 for M, 6C4 for P and 2C2 for
Thus, the required number of ways
= (12C6) (6C4) (2C2)
38. The number of points of intersections
= ( nC2) (2) + (mC2 ) (1) + ( nC1) (mC1 ) (2)
1
= n(n – 1) +
m (m – 1) + 2mn
2
39. Imitate Example 18.
40. Required number of natural numbers
= 9P 1 + 9 ¥ 9P 1 + 9 ¥ 9P 2 + º + 9 ¥ 9P 9
41. When 7 is right in the beginning, the number of
numbers = 94 and when 7 is not in the beginning,
the number of numbers is (4C1) (8) (93)
\ Required number of numbers
= 94 + (4C1) (8) 93 = (41) (93)
7!
5!
= 175
¥
3! 4! 4!
49. The number of ways of putting all the balls in exactly
one box = 5
As for each ball there are 5 choices for the box,
the number of ways of putting balls in the boxes
= 55
Thus, the required number of ways of putting the
balls in the boxes = 55 – 5
50. x 2 + 6x + y2 = 4 fi (x + 3)2 + y 2 = 13
fi (x + 3, y) = (± 3, ± 2) or (± 2, ± 3)
fi x = 0, y = ± 2, x = – 6, y = ± 2,
x = – 1, y = ± 3, y = – 5, y = ± 3
51. Use 20! = 218 38 54 72 (11) (13) (17) (19)
52. When exactly three digits are identical and the remaining two are different, then the number of such
numbers
( 3 C1 ) ÊË 5!ˆ¯ = 60
3!
When three digits are identical and the remaining
two are also identical, then the number of ways
=
5!
( 3 C2 ) (2) ÊÁË 3! 2!ˆ˜¯
= 60
Thus, the number of such numbers = 120.
Permutations and Combinations 6.33
53. Use the fact that r! is divisible by 7 whenever
r ≥ 7.
m
m
n
54. As the last digit of 6 , m ΠN is 6, 6 + 9 will be
divisible by 5 if the unit’s digit of 9n is 4 or 9. This
is possible when n is odd.
\ required number of ordered pairs = 50 ¥ 25 =
1250.
55. The required number of words = 105 –
10
P5
56. To get a square of size r ¥ r, we must choose (r + 1)
consecutive horizontal and (r + 1) consecutive vertical
lines. This can be done in (9 – r) (9 – r) = (9 – r) 2
ways.
57. n C2 = 153
fi
n = 18
58. As groups are indistinguishable, the number of ways
1 mn
=
( Cm) (mn – mCm ) º ( mCm)
n!
=
( mn )!
1
m!
(mn - m )!
n! m ! ( mn - m )! m ! ( mn - 2m )! m ! 0!
1 ( mn )!
=
n! ( m !)n
59. The sequence of six tails may begin at with the ith
toss where i = 1, 2, 3, 4, 5. Thus, the number of ways
is
24 + 1(23) + 2(1) (22) + 22(1)2 + 23(1) = 48
64. For each book we have p + 1 choices.
65. Find the number of positive integral solutions of
x + y + z = 11
where 1 £ x, y, z £ 6.
66. The total number of points is 15. From these 15
points we can obtain 15C3 triangles. However, if all
the 3 points are chosen on the same straight line, we
do not get a triangle. Therefore, the required number
of triangles
=
Thus, the number of onto functions = 34 – 3 – 3(24
– 2) = 36.
C3 – 3(5C3) = 425.
67. The required number of ways
= The number of ways of permuting nine digits out
of ten digits 0, 1, 2, …, 9
– The number of ways of permuting nine digits out
of nine digits 1, 2, …, 9
=
10
P9 - 9 P9 =
10!
- 9! = 10! - 9!
1!
68. The number will be divisible by 25 if it ends in 25
or 75. Therefore, the required number of numbers is
( 5 P2 ) (2) = 40
69. We must have 15 – x ≥ 1 and x – 8 ≥ 0 and 15 – x
≥ x – 8 fi 8 £ x £ 11. Now,
f(8) = 7P0 = 1, f(9) = 6P1 = 6,
f(10) = 5P2 = 20, f(11) = 4P3 = 24,
60. Total number of functions = 34.
All the four elements can be mapped to exactly one element in 3 ways, and exactly two elements in 3(24 – 2).
15
Thus, range of f is {1, 6, 20, 24}.
70. We can arrange remaining 17 boys in 17! ways. For 3
particular boys we have 18 positions. We can arrange
3 particulars boys at these places in 18P3 ways. Thus,
the required number of ways
61. We have
Thus,
C 2 + 3C 2 + 4C 2 + º + n C 2
= ( 3C 3 + 3C 2) + 4C 2 + º + nC 2
= 4C 3 + 4C 2 + º + nC 2
= 5C 3 + º + n C 2 = º = n + 1C 3
x = n + 1C2 + 2( n + 1C3)
= n + 1C2 + n + 1C3 + n + 1C3 = n + 2C3 + n + 1C3
1
n(n + 1) (2n + 1)
=
6
= (17!)
2
71. For each book we have four choices. We can choose
0, 1, 2 or 3 volumes of the book.
72. We have
an
10 n ( n + 1)! n + 1
¥ n +1 =
=
.
an + 1
n!
10
10
Note that
a1
< 1,
a2
62. We have to arrange 12 persons at 14 seats.
63. Let x1 = x + 5, y1 = y + 5, z1 = z + 5 and find the
number of non-negative integral solutions of
x1 + y1 + z1 = 15
18!
(18 P3 ) = 17! 15!
…
Thus,
\
a2
<1,
a3
a3
< 1,
a4
a4
< 1,
a5
a8
a
a
a
< 1, 9 = 1 , 10 > 1, 11 > 1, …
a9
a10
a12
a11
a1 < a2 < a3 < … <a9 = a10 > a11 > a12
an is greatest if n = 9 or 10.
6.34
Complete Mathematics—JEE Main
Ê 2 nˆ
ÁË r ˜¯
(2n)! (r + 1)!(2n - r - 1)!
73. We have
=
r ! (2n - r )!
(2n)!
Ê 2n ˆ
ÁË r + 1˜¯
=
r +1
2n – r
Since for 0 £ r £ n – 1,
r +1
< 1, we get
2n – r
1
(211 – 1) (33 – 1) (54 – 1) (73 – 1) (113 – 1)
480
=
A divisors of m is of the form p1b1 p2 b2 pr br where
0 £ bi £ ai for i = 1, 2, º, r.
That is, bi can take ai + 1 values. Thus, the number
of divisors of m is (a1 + 1) (a2 + 1) º (ar + 1)
76. For the truth of statement-2, see theory of this chapter.
We have
Ê 1000ˆ
1000!
ÁË 500 ˜¯ =
500! 500!
Ê 2 nˆ Ê 2 nˆ
Ê 2 n ˆ Ê 2 nˆ
ÁË 0 ˜¯ < ÁË 1 ˜¯ < < ÁË n – 1˜¯ < ÁË n ˜¯
Ê 2 nˆ Ê 2 n ˆ
Also, as Á ˜ = Á
Ë r ¯ Ë 2n – r ˜¯
Ê 2 nˆ Ê 2 n ˆ º Ê 2 n ˆ Ê 2 n ˆ
<
< Á
ÁË 2n˜¯ < ÁË 2n – 1˜¯
Ë 2n + 1˜¯ ÁË n ˜¯
Ê 2nˆ
Thus, Á ˜ is maximum when r = n.
Ë r¯
Next,
Ê 40ˆ Ê 60ˆ Ê 40 ˆ Ê 60ˆ
ÁË r ˜¯ ÁË 0 ˜¯ + ÁË r – 1˜¯ ÁË 1 ˜¯ + = the number ways of selecting r persons out of 40
men and 60 women
Ê 100ˆ
= Á
Ë r ˜¯
which is maximum when r = 50.
74. The number of ways of distributing n identical objects
among r persons giving zero of more objects to a
person is equivalent to arranging n identical objects of
one kind and (r – 1) identical objects of second kind
(n + r – 1)! Ê n + r – 1ˆ .
in a row, which is equal to
=
n! (r – 1)! ÁË r – 1 ˜¯
Next, the number of non-negative integral solutions of
x1 + x2 + º + x20 = 100 equals the number of ways
of distributing 100 identical objects among 20 persons
giving zero or more objects to a person, which equal
Ê 100 + 20 – 1ˆ Ê 109ˆ .
ÁË 20 – 1 ˜¯ = ÁË 19 ˜¯
75. Sum of the divisors of n
= (1 + 2 + º – 210) (1 + 3 + 32) (1 + 5 + 52 + 53)
(1 + 7 + 72) (1 + 11 + 112)
Ê 3
ˆÊ 4
ˆÊ 3
ˆÊ 3
ˆ
= (211 – 1) 3 – 1 5 – 1 7 – 1 11 – 1
ÁË 3 – 1 ˜¯ ÁË 5 – 1 ˜¯ ÁË 7 – 1 ˜¯ ÁË 11 – 1 ˜¯
Exponent of 11 in the prime factorization of 1000! is
È1000 ˘ È1000 ˘
ÍÎ 11 ˙˚ + ÍÎ 112 ˙˚ = 90 + 8 = 98
Exponent of 11 in the prime factorization of 500! is
È 500 ˘ È 500 ˘
ÍÎ 11 ˙˚ + ÍÎ 112 ˙˚ = 45 + 4 = 49.
Thus, exponent of 11 in the prime factorization of
Ê 1000ˆ
ÁË 500 ˜¯ is 0.
fi
Ê 1000ˆ
ÁË 500 ˜¯ is not divisible by 11.
77. Let
Ê 2n + 1ˆ Ê 2 n + 1ˆ
Ê 2 n + 1ˆ
+Á
++ Á
S= Á
˜
˜
Ë 0 ¯ Ë 1 ¯
Ë n ˜¯
(1)
Ê nˆ
Ê n ˆ
Using Á ˜ = Á
, we can write (1) as
Ë r¯
Ë n – r ˜¯
Ê 2n + 1ˆ Ê 2 n + 1ˆ
Ê 2 n + 1ˆ
++ Á
+Á
S= Á
˜
˜
Ë 2n + 1¯ Ë 2 n ¯
Ë n + 1 ˜¯
(2)
Adding (1) and (2), we get
Ê 2n + 1ˆ
Ê 2 n + 1ˆ
++ Á
= 22 n + 1
2S = Á
Ë 0 ˜¯
Ë 2 n + 1˜¯
fi
S = 22n = 4n
The number of ways of choosing r books out of
Ê 2n + 1ˆ
. We are given
2n + 1 is Á
Ë r ˜¯
fi
Ê 2n + 1ˆ Ê 2n + 1ˆ
Ê 2n + 1ˆ
ÁË 1 ˜¯ + ÁË 2 ˜¯ + + ÁË n ˜¯ = 255
fi
4n – 1 = 255 fi 4n = 256 = 44 fi n = 4.
Permutations and Combinations 6.35
Level 2
k
Now,
78. Arrange m white and n red counters on one side of
(m + n)!
the central mark. This can be done in
m! n !
(2n + 1)(2 n + 2) (4n - 1)(4n)
79. x =
(2 n + 2)(2 n + 4) (4 n)
j =1
81. For the first place we have nine choices. For each of
the next four places, we have 10 choices. At this stage
we add the numbers already selected and choose the
digit at unit’s place in 5 ways.
82. Ak is obtained by putting x = – k everywhere in the
expression except in x + k. Therfore,
m!
Ak =
(- k ) (- k + k - 1)(- k + k + 1) (- k + m)
( m Ck )
84. For each x ΠA, we have four choices;
(i) x ŒP, x ŒQ
= (1 + 2 +
r ! ( n + 1 - r )!
n!
.
(r - 1)! (n - r + 1)! (n + 1)!
r
=
n +1
Also, k2 – 8 =
Thus,
1
£ k2 - 8 £ 1
n +1
1
+ 8 £ k2 £ 9
n +1
fi 8 < k2 £ 9 fi – 3 £ k < –2 2 or 2 2 < k £ 3
Hence, k Œ [–3, – 2 2 )
Previous Years’ AIEEE/JEE Main Questions
1. We can arrange 6 men around the round table in 5!
ways. There are 6 places for women to sit. Out of
these we choose 5 and arrange women there. This
can be done in (5!) (6C5) (5!) = (5!) (6!) ways.
M
W
(iii) x œP, x ŒQ
W
M
(iv) x œP, x œQ
M
W
W
Out of these four choices, last 3 choices imply
x œ P « Q.
Thus, P and Q can be chosen in 3n ways, so that
P « Q = f.
85. As 1! + 2! + 3! + 4! = 33 and last digit of n!, for
+ 2009!)500 is
n ≥ 5 is 0, last digit of (1! + 2! +
500
same as that of last digit of 3
= (10 – 1)
86. Note that d n ¤
+ 2100) (1 + 3 + 32)
87. We have, r – 1 ≥ 0, r £ n + 1
1
r
£
£1
fi1£r£n+1fi
n +1
n +1
(ii) x ŒP, x œQ
But 3 = 9
1)250 is 1.
, 3(2100), 32 (32)(2),
k
Âdj
fi8<
83. As 352706 is even, either both x and y are even or
both x and y are odd.
If x, y are both even, x2 – y2 is divisble by 4, but
352706 is not divisible by 4.
If x, y are both odd, x2 – y2 is divisible by 4, but
352706 is not divisible by 4.
250
n
1 k
Âd
n j =1 j
Ê 2101 - 1ˆ
= Á
(13) = 13(2101 - 1)
Ë 2 - 1 ˜¯
= ( pC 2) ( qC 2) + ( qC 2) ( rC 2) + ( rC 2) ( pC 2)
500
=
j =1
80. Number of parallelograms
(-1)k m!
= (-1)k
k !(m - k )!
j =1
dj
1, 2, , 2100, 3, (3) (2),
(32) (2100)
Thus, x – y + 2n = 2n
=
j
k
=Â
But divisors of n = 2100 32 are
Thus,
1 (4n)! n!
=
=y
(2n)! 2 n (2n)!
1
Âd
n
n
d
250
, and last digit of (10 –
M
M
W
M
W
2. Required number of ways
= (5C4) (8C6) + (5C5) (8C5) = 196
3. nCr+1 + nCr–1 + 2(nCr)
= (nCr–1 + nCr) + (nCr + nCr+1)
=
n+1
Cr +
n+1
Cr+1 =
n+2
Cr+1
Complete Mathematics—JEE Main
6.36
4. We must have
=
7 – x ≥ 1, x – 3 ≥ 0 and 7 – x ≥ x – 3
fi x £ 6, x ≥ 3 and x £ 5
Then, x1 + x2 + x3 + x4 + x5 = 6
\ Range = {4P0, 3P1, 2P2}
= {1, 3, 2}
So the required number of ways is (1/2) 6! = 360.
6. If x1 is the number of balls in the ith box, then we
must have
x1 + x2 + x3 = 8
(1)
The number of solutions of (1) is 7C2 = 21.
n
7. We have tn = Â
r =0
n
r
n
= Â
Cr
r =0
n-r
n
Cr
= ( C4 +
= (51C4 +
51
51
C3 + … +
C 3) +
51
C 3) +
52
55
=…=
56
52
C3 + … +
C3
C3 + … +
55
C3
C3 + … +
55
x1 + x2 + x3 + x4 = 10
n+1
C r]
C3
word beginning with S, therefore, its rank is 601.
10. Required number of ways
=
C1 +
C2 +
(1)
where xi ≥ 1.
Let xi = yi + 1 where yi ≥ 0
Thus, (1) becomes
C4
10
C 3 – 6C 3
16. Let xi = number of balls put in the ith box, then
9. Number of words which begin with A, C, H, I or N
is 5(5!) = 600. These have ranks from 1 to 600. Then
10
10
1
1
(10 ¥ 9 ¥ 8) –
(6 ¥ 5 ¥ 4)
6
6
= 100
C 3)
55
[using nCr + nCr–1 =
C4 +
C4 = N.
=
C4 + (50C3 +
50
52
10
Thus, statement-2 is true.
N=
r =1
=
10!
=
4 !6 !
15. Number of triangles
tn
n
= .
sn
2
6
50
=
= (3C2) (9C2) = (3) (36) = 108
8. 50C4 + Â 56- r C3
=
The number of ways of arranging 6 A’s and 4 B’s in
a row
= (6C4) (3C1) (4!) = 1080 > 1000
Cn - r
= n sn – tn
fi 2tn = n sn fi
50
\ Statement-1 is false.
14. Number of ways
n-r
n
The number of non-negative integral solution of (1)
C5–1 = 10C4 = N (say)
13. We can choose 4 novels out of 6 in 6C4 ways and
1 dictionary out of 3 in 3C1 ways. We can arrange
4 novels and 1 dictionary in the middle in 4! Ways.
Thus, required number of ways
where xi ≥ 1.
r =0
(1)
6+5–1
5. We can arrange the letters of the word GARDEN in
6! ways. Exactly half of these will be with vowels in
alphabetical order.
n
(41)3
12. Let xi = number of ice creams of ith type bought by
the child.
Thus, 3 £ x £ 5
= Â
12!
10
C3 +
10
C4
= (12C4) (8C4) (4C4)
(2)
Number of non-negative integral solutions of (2) =
number of ways of arranging 6 identical balls and 3
identical separators in a row.
=
= 10 + 45 + 120 + 210 = 385
11. Number of ways
y1 + y2 + y3 + y4 = 6
(6 + 3)! 9
= C3
6!3!
= number of ways of choosing 3 places out of 9 different places.
17. We can choose 0 or more balls out of 10 white balls
in 11 ways.
Permutations and Combinations 6.37
[Select 0 or 1 or 2, … or 10 balls.]
A
Thus, the number ways of selecting 0 or more balls
is (1 + 10) (1 + 9) (1 + 7) = 880.
Therefore, the number of ways selecting one or more
balls is 880 – 1 = 879.
18. As Tn = nC3,
Tn+1 – Tn = 10
fi n+1C3 – nC3 = 10
fi nC2 + nC3 – nC3 = 10
fi
( 3C 2) ( 9C 1) + ( 4C 2) ( 8C 1) + ( 5C 2) ( 7C 1)
= 27 + 48 + 70 = 145.
Thus, number of required triangle is 60 + 145 = 205.
23. Each digit occurs at the unit’s place exactly 3! = 6
times. This, sum of the digits at the units place
19. Let xi = marks assigned to ith question, then
x1 + x2 + … + x8 = 30,
= 6[3 + 4 + 5 + 6] = 108
xi ≥ 2 for i = 1, 2, …, 8.
Put xi = yi + 2, then
y1 + y2 + … + y8 = 14
(1)
The number of solutions of (1) is
C8–1 =
21
24. As 0 + 1 + 2 + … + 9 = 45, a number will be divisible by 9, if we do not use (0, 9), (1, 8), (2, 7), (3,
6), (4, 5).
Thus, the number of 8 digit number that are divisible
by 9 is 8! + (8! – 7!) (4) = (36) (7!)
C7
20. Any number that is formed by 2, 3, 5, 7, 9 exceeds
20000.
25. We can choose places for 1, 1, 3 in 4C3 = 4 ways.
\ We can arrange the digits
\ p = 5!
1, 1, 2, 2, 2, 3, 4, 4 is
A number that begins with 3, 5 or 7 lies between
3000 and 90000. Therefore,
(4)
q = (3) (4!)
3! 5!
= 120
2! 3!2!
26. Number of games played by men among themselves
= 2(nC2) = n(n – 1)
Thus, p : q = 5 : 3
21. Different number of ways of forming committee are
as under.
Ladies
Old men
Young men
OPTION 1
1
1
2
OPTION 2
1
2
1
OPTION 3
2
1
1
Thus, the required number of ways
Number of games played by men with women = 2(2n)
= 4n
\ n(n – 1) – 4n = 66 fi n(n – 5) = 66
fi (n Р11) (n + 6) = 0 fi n = 11 Π[10, 12)
27. n(A ¥ B) = n(A) n(B) = (4) (2) = 8
\ number of subset of A ¥ B which contain at least
three elements
= 8C 3 + 8C 4 + 8C 5 + 8C 6 + 8C 7
= ( 2C 1) ( 2C 1) ( 4C 2) + ( 2C 1) ( 2C 2) ( 4C 1) +
( 2C 2) ( 2C 1) ( 4C 1)
= 2 8 – ( 8C 0 + 8C 1 + 8C 2)
= 24 + 8 + 8 = 40.
= 256 – (1 + 8 + 28) = 219
22. Take one vertex on each side.
This can be done in (3C1) (4C1) (5C1)
= 60 ways.
C
Choose two vertices on one side and one from the
remaining two sides. This can be done in
1
n(n – 1) = 10 fi n = 5
2
14+8–1
B
28. Any four digit number beginning with 6, 7 or 8 is
greater than 6000. Thus, required number of numbers
is
(3P1) (4P3) + 5P5 = 192
Complete Mathematics—JEE Main
6.38
Last word (60th word) in the list is SMLLA, 59th word
is SMLAL, and 58th word is SMALL.
29. For 1 £ k £ 40, the number of integral points on the
line x = k and lying in the interior of triangles is
k – 1.
Alternate solution
40
Number of words beginning with
4!
(i) A is
= 12
(Ranked 1 to 12)
2!
Thus, required number of points = Â (k - 1)
k =1
1
(39) (40) = 780
2
(0, 41)
k)
(k, k - 1)
(k,
(0, 0)
1
(k, 1)
k
40
(ii) L is 4! = 24
4!
= 12
(iii) M is
2!
(iv) SA is
(41, 0)
30. If we allow same man (woman) to be in two distinct
teams (for instance if (M1, W1) and (M1, W2) are two
distinct teams), then the answer is 225C15.
3!
=3
2!
(v) SL is 3! = 6
(vi) SM (ALL) is 1
15
33.
15
Cr
Cr -1
=
(Ranked 13 to 36)
(Ranked 37 to 48)
(Ranked 49 to 51)
(Ranked 52 to 57)
(Ranked 58th)
15!
(r - 1)!(16 - r )!
◊
r !(15 - r )!
15!
15 - r
r
If we do not allow same man (woman) to be in two
=
(15) ways, second team can be chosen in (14) (14)
ways and so on.
Ê 15 Cr ˆ
\ ar = r Á 15
˜ = r(15 – r)
Ë Cr -1 ¯
As all the teams 15 are to be chosen, the number of
ways
= (15)2 (14)2 (13)2 … (22) (12)
2
15
15
r =1
r =1
fi  ar =  r (16 - r )
= (15!)2
¸ 1
Ï1
= 16 Ì (15)(16) ˝ - (15)(16)(31)
2
Ó
˛ 6
It seems that answer 1240 is arrived by the following
way
=
1
31˘
È
(15)(16) Í16 - ˙
3˚
2
Î
=
1
Ê 17 ˆ
(15)(16) Á ˜ = 680
Ë 3¯
2
152 + 142 + … + 22 + 12
1
(15) (16) (2 ¥ 15 + 1) = 120
6
This is an incorrect way as we have to select all the
15 teams.
=
31. Let number of sides of the regular polygon be n. The
number of diagonals of the polygon
= nC 2 – n =
Thus,
1
1
n(n – 1) – n = n(n – 3)
2
2
1
n(n – 3) = 54
2
fi n2 – 3n – 108 = 0
fi (n – 12) (n + 9) = 0
As n ≥ 3, n = 12.
32. Number of 5 letter words that can be formed by using
5!
letters of word SMALL is
= 60.
2!
34. As r2 + 1 = (r + 2) (r + 1) – 3(r + 1) + 2, we get
(r2 + 1)r! = (r + 2)! – 3(r + 1)! + 2(r!)
= [(r + 2)! – (r + 1)!] – 2[(r + 1)! – r!]
10
fi  (r 2 + 1)r ! = (12! – 2!) – 2(11! – 1!)
r =1
= 11!(12 – 2) = (10)(11!)
35. Letters in the word M E D I T E R R A N E A N are
A, A, D, E, E, E, I, M, N, N, R, R, T. Two middle
letters must be chosen from
A, A, D, E, E, I, M, N, N, R, T. If the chosen letters
are distinct, then these can be arranged in 8P2 = 56
ways.
If the chosen letters are indentical, then these can be
chosen and arranged in 3C2 = 3 ways.
Thus, total number of such letters is 59.
Permutations and Combinations 6.39
36.
W1 M1 W2 M2 … Wn Mn
(n + 2)! (n - 4)!
◊
= 11
6!(n - 4)! (n - 2)!
fi
This can be done in (2) (n!) (n!) = 2(n!)2 ways.
(n + 2)(n + 1)(n)(n - 1)
= 11
(6)(5)(4)(3)(2)
6. Number of points of intersections
= (8C2) + 2(4C2) + 2(8C1) (4C1)
fi (n + 2) (n + 1) (n) (n – 1) = (11) (10) (9) (8)
fi n + 2 = 11 fi n = 9
7. Required number of ways
fi n2 + 3n – 108 = 0
Previous Years’ B-Architecture Entrance
Examination Questions
1. The number of required subsets is
N=
2007
C0 +
2007
N=
2007
C2007 +
2007
C1 +
2007
C2 + … +
C2006 + … +
2007
2007
C1003
C1004
Adding above two equations we get
2N =
2007
fiN=2
C0 +
2007
= 104
C1 + … +
2007
= (26 – 1) (26 – 1) – 36 = 212 – 27 – 35
8. If A and B
f : A Æ B is an
injective mapping, then n(A) £ n(B).
As m > n, there does not exist any mapping from set
with m elements to a set with n elements.
9. 10Cx–1 > 2(10Cx)
C2007
fi
2006
2. There are (n + 2) line segments from A to B and
(n + 2) line segments from A to D.
D
= ( 6C 1 + 6C 2 + … + 6C 6) ( 6C 1 + 6C 2 + … + 6C 6) –
( 6C 1) ( 6C 1)
C
10!
10!
>2
x !(10 - x)!
( x - 1)!(11 - x)!
fi x > 2(11 – x) fi 3x > 22
fi x > 7 fi x ≥ 8.
Thus, least positive integral value of x is 8.
10. Number of ways of arranging beds
= 6P6 = 6! = 720
A
Number of ways in which Madhu and Puja have their
beds together is (2) (5P5) = 2(5!) = 240.
B
To get a parallelogram, choose two line segments from
A to B and two line segments from A to D.
\ number of required ways
=(
n+2
C 2) (
n+2
C 2)
Then number of onto mappings
= 2n – 1 – 1 = 2n – 2
4. Required number of subsets
C4 +
10
C5 + … +
= 210 – (10C0 +
10
10
C1 +
C10
10
C2 +
10
C 3)
= 1024 – (1 + 10 + 45 + 120) = 848
5. There are following ways for the photograph.
M1 W1 M2 W2 … Mn Wn
or
Case 1: Both the drivers are occupying the car. In this
case we have to choose 3 persons out of 5 for the
remaining seats and car can be driven by any of the
two drivers. Thus, in this case car can be occupied
and driven in (5C3) (2C1) (4!) = 480 ways.
Case 2: Exactly one drive is occupying the car. In this
case we have to choose 4 persons out of the remaining
5 people. In this case car can be occupied and driven
in (5C4) (2C1) (4!) = 240 ways.
Thus, 2n – 2 = 30 fi n = 5.
10
= 720 – 240 = 480.
11. Two cases arise:
3. Suppose the set contains n elements.
=
Thus, required number of ways
Thus, the required number of ways
= 480 + 240 = 720
n
12. Cr–1 + nCr =
n
\
n
n+1
Cr -1
n
Cr -1 + Cr
Cn
=
n!
r !(n + 1 - r )!
◊
(r - 1)!(n - r + 1)!
(n + 1)!
Complete Mathematics—JEE Main
6.40
=
fi 13n2 = 144(n + 1) fi n = 12, –12/13
r
n +1
n
Ê r ˆ
fi S = ÂÁ
˜
r =1 Ë n + 1¯
n2
36
=
=
4(n + 1)
13
Thus, n = 12
3
=
2
1
(n + 1)
3
◊
n (n + 1)
4
2
13. Number of code words ending with even digit
= (21) (20) (9) (4)
\ 432k = (21) (20) (9) (4)
fi k = 35.
CHAPTER SEVEN
4. The term nCr xn – r yr is the (r + 1)th term from the
beginning of the expansion. It is usually denoted by
Tr + 1 and is called the general term of the expansion.
PRINCIPLE OF MATHEMATICAL
INDUCTION
Let P(n) be a statement involving the natural number n. To
prove that P(n) is true for all natural numbers n, we proceed
as follows:
1. Prove that P(1) is true.
2. Assume that P(k) is true.
3. Using steps 1 and 2, prove that P(k + 1) is true.
MIDDLE TERM(S)
1. n is even
If n is even, then the expansion (1) has just one middle term,
n
viz. ÊÁ + 1ˆ˜ th term. It is given by
Ë2 ¯
n
Note
Sometimes step 2 is replaced by step 2¢.
2¢. Assume that P(n) is true for all n Œ N, where
1 £ n £ k.
The First Principle of Mathematical Induction comprises
steps 1, 2 and 3. Steps 1, 2¢ and 3 constitute another principle
of mathematical induction, usually called the Second
Principle of Mathematical Induction.
BINOMIAL THEOREM (FOR A
POSITIVE INTEGRAL INDEX)
If n is a positive integer and x, y are two complex numbers,
then
(x + y)n = nC0 xn + nC1 xn – 1 y + nC2 xn – 2y2 + . . .
+ nCr xn – r yr +…+ nCn – 1 x yn – 1 + nCnyn
(1)
n
n
n
The coefficients C0, C1,
, Cn are called binomial
coefficients.
PROPERTIES OF THE BINOMIAL
EXPANSION
1. There are (n + 1) terms in the expansion (1).
2. In any term of (1), the sum of the exponents of x
and y is always n.
3. Binomial coefficients of terms equidistant from the
beginning and the end are equal, since nCr = nCn – r
(0 £ r £ n).
Cn/2 xn/2 yn/ 2.
2. n is odd
If n is odd, then the expansion (1) has two middle terms, viz.
Ê n+1 ˆ
Ê n + 1ˆ
+ 1˜ th term. These are given by
ÁË
˜ th term and ÁË
¯
2 ¯
2
and
Illustration
n
C( n -1) 2 x ( n -1) 2 y( n +1) 2
n
C( n +1) 2 x ( n +1) 2 y( n -1) 2
1
3
2 10
Find the middle term of Ê x - yˆ
Ë2
5 ¯
As n = 10 is even, middle term of
2 ˆ 10
Ê10 ˆ
Ê3
x
y
is Ë + 1¯ th or 6th term.
Ë2
5 ¯
2
It is given by
10 - 5
T6 = T5 + 1 =
10
Ê3 ˆ
C5 Ë x ¯
2
5
Ê 2 ˆ
y
Ë 5 ¯
5
Ê 3ˆ
= –252 Ë ¯ x5 y5
5
Illustration
2
2 ˆ 11
Ê1
Find the middle terms of Ë x +
x
2
3 ¯
7.2
Complete Mathematics—JEE Main
As n = 11, is odd, the binomial expansion of
2 ˆ 11
Ê11 + 1 ˆ
Ê1
x
+
x
has two middle terms viz. Ë
th
Ë2
3 ¯
2 ¯
Ê11 + 1 ˆ
and Ë
+ 1¯ th terms, that is, 6th and 7th terms.
2
11 - 5
5
Ê1 ˆ
Ê2 ˆ
T6 = T5 + 1 = 11C5 Ë x ¯
x
Ë3 ¯
2
=
11! 1 25 6+5/2
◊
◊ x
5! 6! 26 35
=
77 17 2
x
81
and T7 = T6 + 1 =
=
11
Ê1 ˆ
C6 Ë x ¯
2
11- 6
Ê2 ˆ
Ë 3 x¯
r
Tr +1
7! (r - 1)! (8 - r )! 57 - r ( -3 x )
=
Tr
r ! (7 - r )! 7! 57 - r + 1 ( - 3 x )r -1
8-r 1
=
(–3x)
r 5
For x = 2/3,
Tr +1
(8 - r ) (2)
=
Tr
5r
Tr + 2 (7 - r ) (2 )
Similarly
=
Tr + 1
5 (r + 1)
For (r + 1)th term to be numerically greatest,
|Tr| < |Tr + 1| and |Tr + 2| < |Tr + 1|
fi 5r < 16 – 2r and 14 – 2r < 5r + 5
fi 9 < 7r < 16 fi r = 2.
Thus, 3rd term is numerically greatest.
fi
6
11! 1 26 5 + 3
308 8
◊ 5◊ 6x
=
x
243
5! 6! 2 3
The Greatest Coefficient
If n is even, the greatest coefficient in the expansion of
(x + y)n is nCn/2.
If n is odd, there are two greatest coefficients in the expansion of (x + y)n. These are nC(n – 1)/2 and nC(n + 1)/2.
Some Other Useful Expansions
In the following expansions Cr stands for nCr
1. (x + y)n + (x – y) n
= 2 [C0 x n + C2 x n – 2 y2 + C4 x n – 4 y4 + . . . .]
2. (x + y)n – (x – y)n
= 2 [C1 xn – 1 y + C3 xn – 3 y3 + C5 xn – 5 y5 + . . . .]
3. n (n–1) (n –2)… (n– k + 1) (x + y) n–k
n
The Greatest Term
To find the greatest term in the expansion of (x + y)n, x, y
> 0, n ΠN put
(n + 1) y
x+y
If a is not an integer, then the expansion of (x + y)n has
just one greatest term and it is given by k = [a] where [a]
denotes the greatest integer < a.
If a is an integer, then expansion of (x + y)n has two greatest
terms viz. ath and (a + 1)th terms.
a=
r =k
4. (1 + x)n + (1 – x)n = 2 [C0 + C2 x2 + C4 x4 + . . . .]
5. (1 + x)n – (1 – x)n = 2 [C1 x + C3 x3 + . . . .]
6. C1 + 2C2 x + 3C3 x2 + . . . . + nCn xn – 1
= n(1 + x)n – 1
x2
x3
x n +1
7. C0 x + C1
+ C2
+ .... + Cn
2
3
n +1
x
= Ú (1 + t )n dt
0
SOME PROPERTIES OF THE BINOMIAL
COEFFICIENTS
TIP
If we are given that tr is numerically the greatest term, we
use
|tr–1| < | tr |
and
|tr+1| < | tr |
to obtain two inequalities from which some other results
can be obtained.
Illustration
= Â r (r - 1) (r - 2)º(r - k + 1)Cr x r - k y n -r
3
Find numerically greatest term in the expansion of
(5 – 3x)7 when x = 2/3.
We have
Tr + 1 = 7Cr 57 – r (–3x)r
C0 + C1 + C2 + . . . . + Cn = 2n
C0 + C2 + C4 + . . . . = C1 + C3 + C5 + . . . . = 2n – 1
C0 – C1 + C2 – . . . . + (–1)n Cn = 0
For n > 1,
C1 – 2C2 + 3C3 – . . . . + (–1)n – 1 nCn = 0
Pn + 1 ( n + 1)n
5. If Pn = C0 C1 C2 . . . . Cn, then
=
Pn
n!
1.
2.
3.
4.
REVERSING TECHNIQUE
If a0, a1, a2, . . . ., an are in A.P., then sum of the series
a0 C0 + a1 C1 + . . . . + an Cn
can be obtained by the reversing technique explained as
follows.
Mathematical Induction and Binomial Theorem
Let
S = a0 C0 + a1 C1 + a2 C2 + . . . . + an – 1 Cn – 1 + an Cn (1)
Using Cr = Cn – r and reversing the order in which terms are
written above, we obtain
S = an C0 + an – 1 C1 + an – 2 C2 + . . . . + a1 Cn – 1 + a0 Cn
(2)
Adding (1) and (2) we get
2S = (a0 + an) C0 + (a1 + an – 1) C1 + (a2 + an – 2) C2 + . . . . +
(an – 1 + a1) Cn – 1 + (an + a0) Cn
As a0, a1, a2, …an are in A.P., we have
a0 + an = a1 + an – 1 = a2 + an – 2 = . . . .,
2S = (a0 + an) (C0 + C1 + C2 + . . . . + Cn)
= (a0 + an)2n.
1
fi
S=
(a0 + an)2n = (a0 + an)2n – 1.
2
e.g. (i) C0 + 2C1 + 3C2 + . . . . + (n + 1)Cn
so that
= (1 + n + 1)2n – 1 = (n + 2)2n – 1
(ii) C1 + 2C2 + . . . . + nCn
stance, to find the remainder when 7200 is divided
by 50, we write 72 = 50 – 1 and then use binomial
theorem.
2. If a, b, r ΠQ and r is an irrational number, then sometime it is useful to switch from
a + b r to a – b r .
3. If p, q ΠQ and
p,
q are irrational, then often
it is preferable to simplify ( p + q )2n by first
using ( p + q )2 = p + q+ 2 pq .
4. If three consecutive binomial coefficients nCr – 1,
1
n
Cr, nCr + 1 are in A.P., then r = (n + n + 2 )
2
5. Four consecutive binomial coefficients can never be
in A.P.
6. Three consecutive binomial coefficients can never
be in G.P. or H.P.
SOME PARTICULAR EXPANSIONS
= 0C0 + 1C1 + 2C2 + . . . . + nCn
= (0 + n)2n – 1 = n(2n – 1)
MULTINOMIAL THEOREM
If we wish to expand (x + y + z)n (where n is a positive integer and x, y, z are complex numbers), we may take y + z =
a, expand the binomial (x + a)n, and then expand (y + z)r in
each term of the expansion of (x + a)n. In general, we have
the following result:
n!
n1 n2
nm
(a1 + a2 + ...+ am)n = Â n ! n ! n ! a1 a2 am
(1)
1
2
m
where the summation is taken over all non-negative integers
n1, n2, ..., nm such that n1 + n2 +
+ nm = n.
The number of distinct terms in the expansion (1) is
n+m–1
C m – 1.
SOME USEFUL TIPS AND TRICKS
1. To find remainder when x n is divided by y, try to
express x or some power of x as ky ± 1. For in-
For | x | < 1
1. (1 + x)–1 = 1 – x + x2 – x3 + x4 – . . . .
2. (1 – x)–1 = 1 + x + x2 + x3 + x4 + . . . .
3. (1 + x)–2 = 1 – 2x + 3x2 – 4x3 + 5x4 – . . . .
4. (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + 5x4 + . . . .
5. (1 + x)–3 = 1 – 3x + 6x2 – 10x3 + 15x4 – . . . .
6. (1 – x)–3 = 1 + 3x + 6x2 + 10x3 + 15x4 + . . . .
7. (1 – x)–n = 1 + nC1 x + n +1C2 x2 + n +2C3 x3 + . . . .
where if n is a natural number.
Expansion when x is very small
If x is very small so that x2 and higher powers of x can be
neglected, then
(1 + x)n = 1 + nx.
If x is very small so that x3 and higher powers of x can be
neglected, then
1
(1 + x)n = 1 + nx +
n (n –1) x2
2
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Example 1: Suppose a ΠR. If the coefficient of x5 in
1ˆ
Ê
the expansion of Á ax + 3 ˜
Ë
x ¯
7.3
17
is 680, then a is equal to
(a) ± 2
(c) ± 1/2
Ans. (b)
(b) ± 1
(d) ± 1/3
7.4
Complete Mathematics—JEE Main
(a) 1
(c) –3
Ans. (a)
r
Ê 1ˆ
Solution: Tr + 1 = 17Cr (ax)17 – r Á 3 ˜
Ëx ¯
= 17Cr a 17 – r x17 – 4r
For the coefficient of x5, we set
17 – 4r = 5 or r = 3.
17
C3 a14 = 680
\
680 ¥ 3! 14!
fi
a14 =
=1
17!
fi
a = ± 1.
10
Ê10 ˆ
Ê k xˆ
Solution: Middle term of Ë + ¯ is Ë + 1¯ th =
x k
2
6th term. We have
\
Example 2: The absolute term (that is, term independ1 ˆ9
Ê3
ent of x) in the expansion of Ë x 2 - ¯ is:
2
3x
1
2
(a)
(b)
9
17
7
11
(c)
(d)
18
18
Ans. (c)
9-r
Ê3 ˆ
Solution: Tr + 1 = 9Cr Ë x 2 ¯
2
5
Ê xˆ
10
Ë k ¯ = C5
Ê k2 x ˆ
Also, (r + 1)th term of ÁË
+ ˜
x k¯
10 - r
10
Ê xˆ
Ë k¯
is
r
= 10Cr k20 – 3r x2r – 10
For term independent of x, set 2r – 10 = 0 or r = 5. Thus,
10
r
Ê 1ˆ
Ë 3x ¯
9 -r
Solution: T r + 1 = 10Cr (2y)r = (10Cr 2r) y r.
\
ar = 10Cr 2 r
Now
ar + 2 10 Cr + 2 2r + 2
4=
= 10
ar
Cr 2r
10
10 - 5
Ê kˆ
T6 = 10C5 Ë ¯
x
a = 10C5
Ê k2 ˆ
Tr + 1 = 10Cr ÁË ˜¯
x
r
Ê3 ˆ Ê 1ˆ
= 9Cr Ë ¯ Ë - ¯ x18 – 3 r
2
3
For term independent of x, we set 18 – 3r = 0 or r = 6. Thus,
1 ˆ9
Ê3
term independent of x in the expansion of Ë x 2 - ¯ is
2
3x
9-6
6
7
Ê3 ˆ
Ê 1ˆ
9
C6 Ë ¯
- ¯ =
Ë
3
18
2
Example 3: Let ar denote the coefficient of yr – 1 in the
ar + 2
expansion of (1 + 2y)10. If
= 4, then r is equal to
ar
(a) 2
(b) 3
(c) 4
(d) 5
Ans. (c)
fi
(b) 2
(d) any non-zero number
Cr + 2 = 10Cr fi r + 2 = 10 – r fi r = 4
Example 4: Suppose k ΠR. Let a
10
Ê k xˆ
the middle term in the expansion of Ë + ¯ , and b be the
x k
Ê k2 x ˆ
term independent of x in the expansion of ÁË
+ ˜ is
x k¯
10
C5 k5 = b.
a
= 1, we get k5 = 1 or k = 1
As
b
Example 5: The remainder when 6n – 5n is divided by
25, is
(a) 1
(b) 24
(c) 0
(d) n
Ans. (a)
Solution: 6n – 5n = (1 + 5)n – 5n
= 1 + 5n + nC2(52) + nC3(53) + . . . + nCn(5n)
– 5n
= 1 + 25m
where
m = nC2 + nC3(5) + . . . + nCn(5n – 2)
is a whole number.
Thus, remainder when 6n – 5n is divided by 25 is 1.
Example 6: If coefficients of r th and (r + 1)th term
in the expansion of (3 + 2x)74 are equal, then r is equal to:
(a) 28
(b) 29
(c) 30
(d) 31
Ans. (c)
Solution: Tr + 1 = 74Cr 374 – r (2x)r
fi
r + 1)th term is 74Cr 374 – r 2r
We are given
74
Cr – 1 374 – r + 1 2 r – 1 = 74Cr 374 – r 2 r
3
74!
(r - 1)! (75 - r )!
◊
fi
=
2
r ! (74 - r )!
74!
10
Ê k2 x ˆ
+ ˜ . If
term independent of x in the expansion of ÁË
x k¯
a
= 1, then k is equal to
b
75 - r
r
3r = 150 – 2r fi
=
fi
r = 30
Mathematical Induction and Binomial Theorem
Example 7: If the coefficients of three consecutive
terms in the expansion of (1 + x)n are in the ratio 1 : 7 : 42,
then value of n is
(a) 45
(c) 65
Ans. (b)
Solution: Let three consecutive terms be (r + 1)th,
(r + 2)th and (r + 3)th. Then
n
n
Cr
Cr +1
=
Example 9: Sum of the series
8
S=
Â
1
j=0
(b) 55
(d) 75
r +1 1
1
= fi n – 8r = 7
fi
n-r 7
7
( j + 1) ( j + 2 )
and
n
Cr + 1
Cr + 2
=
1003
90
(b)
1013
90
(c)
1023
90
(d)
1033
90
Ans. (b)
(1)
Solution: For 0 £ j £ 8,
( j + 1) ( j + 2 )
1
7 1
r+2
=
= fi
42 6
n - r -1 6
fi
n – 7r = 13
From (1) and (2), n = 55.
( 8 C j ) is
(a)
1
n
Example 8: The coefficient of x in the expansion of E
= (1 + x)9 + (1 + x)10 + . . . + (1 + x)100 is
(b) 101C10
(a) 101C9
100
(c) C9
(d) 100C10
Ans. (b)
(1 + x )9 ÈÎ(1 + x )92 - 1˘˚
Solution: E =
(8C j )
=
8!
( j + 1) ( j + 2 ) j ! (8 - j )!
=
1
10!
10 ¥ 9 ( j + 2 )! (10 - ( j + 2 ))!
=
1
90
(2)
9
( 10 C j + 2 )
Thus,
S=
1+ x -1
ˆ
1 Ê 10 10
Cj + 2˜
Â
Á
90 Ë j = 0
¯
(1 + x )101 - (1 + x )9
=
x
Coefficient of x9 in E
= coefficient of x10 in [(1 + x)101 – (1 + x)9]
= 101C10
ˆ
1 Ê 10 10
Ck - 10C0 - 10C1 ˜
Â
Á
90 Ë k = 0
¯
=
1
1013
(210 – 1 – 10) =
90
90
=
Example 9: If the 7th term in the binomial expansion of
9
ˆ
Ê 3
ÁË 1 3 + 3 ln x˜¯ , (x > 0) is equal to 729, then x is equal to
84
(a) e, 1/e
(c) e2, e–2
Ans. (a)
(b) 2e, 1/2e
(d) 1/e
9-6
Ê 3 ˆ
Solution: T7 = 9C6 Á 1 3 ˜
Ë 84 ¯
(
6
3 ln x )
fi
9 ¥ 8 ¥ 7 33 3
3 (ln x)6 = 729
◊
3 ¥ 2 84
fi
fi
(ln x)6 = 1
ln x = ± 1 fi x = e, 1/e
7.5
Example 10: If sum of the coefficients in the expansion of (x + y)n is 2048, then the greatest coefficient in the
expansion is:
(b) 11C6
(a) 10C6
10
(c) C7
(d) 12C6
Ans. (b)
Solution: Sum of the coefficients in the expansion of (x
+ y)n is 2n. Therefore,
2n = 2048 = 211 fi n = 11.
As n is odd, the greatest value of nCr is nCk where
1
1
k = (n – 1) or k = (n + 1).
2
2
Here
n = 11, therefore k = 5 or 6.
Thus, greatest coefficient is 11C6.
7.6
Complete Mathematics—JEE Main
LEVEL 1
Straight Objective Type Questions
Example 11: 5th term from the end in the expansion of
8
Ê x2
3ˆ
- 2 ˜ is:
ÁË
3 x ¯
is:
(a) 25x2
(c) 70
Ans. (c)
(b) –70
(d) –25x2
(a) 37th, 38th
(c) 47th, 48th
Ans. (b)
Solution: Suppose coefficients of (r + 1)th and (r + 2)th
terms in the expansion of (3 + 2x)94 are equal. Then
94
Cr 394 – r 2r = 94Cr + 1 394 – (r + 1) 2r + 1
(r + 1)! (94 - r - 1)! 2
94!
◊
=
94!
3
r ! (94 - r )!
fi
Solution:
fi
TIP
As number of terms in the binomial expansion of (x + a)n is
(n + 1), the rth term from the end is ((n + 1) + (–1) (r – 1))th
= (n + 2 – r)th term from the beginning.
Here n = 8, r = 5, thus, 5th term from the end is (8 + 2 – 5)th
= 5th term from the beginning, and it is given by
4
Ê x2 ˆ Ê 3 ˆ 8 - 4 8
= C4 = 70
t5 = C4 ÁË ˜¯ Á - 2 ˜
Ë x ¯
3
Example 12: If
8
ÈÊ 1 + 3 x + 1 ˆ n Ê 1 - 3 x + 1 ˆ n ˘
ÍÁ
P(x) =
˜¯ ˙
˜¯ - ÁË
˚
5
5
3 x + 1 ÎË
is a 5th degree polynomial, then value of n is
(a) 9
(b) 11
(c) 23
(d) 21
Ans. (b)
1
Solution: P(x) = n [(1 + a)n – (1 – a)n]
5 a
1
where
Now,
a=
P(x) =
3x + 1
2
n
[ nC 1 a + nC 3 a 3 + . . . ]
5 a
Note that n must be odd, so that
in P(x).
\
P(x) =
2
5n
2
3 x + 1 does not appear
[ nC 1 + nC 3 a 2 + . . . + nC n a n – 1]
[nC1 + nC3 (3x + 1) + nC5 (3x + 1)2
5n
+ . . . + nCn (3x + 1)(n – 1)/2]
n -1
For this to be a polynomial of degree 5,
= 5 fi n = 11.
2
Example 13: The two consecutive terms whose coefficients in the expansion of (3 + 2x)94 are equal, are
=
(b) 38th, 39th
(d) 48th, 49th
fi
2
r +1
=
fi 3r + 3 = 188 – 2r
3
94 - r
5r = 185 fi r = 37
Example 14: Coefficient of the constant term in the
expansion of
-9
1
Ê 1
ˆ
E = (x2/3 + 4x1/3 + 4)5 Á 1 3
+ 23 13 ˜
Ë x - 1 x + x + 1¯
is
(a) 74
(c) 148
Ans. (d)
(b) 98
(d) 168
Solution: E = ((x
1/3
Ê x 2 3 + x1 3 + 1 + x1 3 - 1ˆ
+ 2) ) Á
˜¯
Ë
x -1
-9
2 5
Ê x 2 3 + 2 x1 3 ˆ
= (2 + x1/3)10 Á
Ë
x - 1 ˜¯
= (2 + x1/3)10
-9
( x - 1)9
( x1 3 )9 ( x1 3 + 2 )9
= (2 + x1/3)x–3 (x – 1)9
Constant term in E
= 2[constant term in x–3 (x – 1)9]
= 2(–1)9 [coefficient of x3 in (1 – x)9]
= 2(–1) 9C3 (–1) = 168
Example 15: Coefficient of x4 in the expansion of
2 ˆ6
Ê
x
1
+
3
+
˜ is:
ÁË
x2 ¯
(a) 240
(c) 820
Ans. (b)
(b) 735
(d) 936
Solution: The general term in the expansion of
2 ˆ6
Ê
ÁË 1 + 3 x + 2 ˜¯ is
x
Mathematical Induction and Binomial Theorem
Example 18: The coefficient of x7 in the expansion of
c
6!
Ê 2ˆ
1a (3 x )b Á 2 ˜
Ëx ¯
a ! b! c !
where a, b, c ≥ 0 and a + b + c = 6
1
1 2 1 3 1 4 1 5ˆ2
Ê
1
+
x
+
x + x + x + x ¯
Ë 1!
2!
3!
4!
5!
6! 3b 26 - a - b
xb + 2a + 2b – 12
=
a ! b! (6 - a - b )!
for coefficient of x4, set 2a + 3b – 12 = 4
Now 2a + 3b = 16 and a + b + c = 6 fi (a, b, c) = (2, 4, 0)
Thus, required coefficient is
6! (34 ) (20 )
2! 4!
is:
(a)
2 8
( C3 )
7!
(b)
210
5!
(c)
4
15
(d)
7
20!
Ans. (a)
Solution: Required coefficient is
1
1
1
1
+
+
+
2! 5! 3! 4! 4! 3! 5! 2!
= 735
Example 16: Coefficient of x17 in the polynomial P(x)
17
=
’ ( x + 35Cr ) is
=
1 7
( C 2 + 7C 3 + 7C 4 + 7C 5)
7!
=
2 7
2 8
( C 2 + 7C 3) =
( C 3)
7!
7!
r=0
(a) 234
(c) 235 – 36C17
Ans. (a)
(b) 36C17
(d) 0
Example 19: Sum of the series
Solution: P(x) is a polynomial of degree 18, and coefficient of x17 is
N = 35C0 + 35C1 + 35C2 + . . . + 35C17
(1)
n
n
Using Cr = Cn – r, we can write N as
(2)
N = 35C35 + 35C34 + . . . + 35C18
Adding (1) and (2), we get
2N = 35C0 + 35C1 + . .. + 35C35 = 235
fi
N = 234
•
k
 Â
k =1 r = 0
22r
7k
1
1 2
1 20 ˆ 3
Ê
1
+
x
+
x
+
.
.
.
+
x ¯
Ë 1!
2!
20!
is:
(kCr) is:
(a) 1/7
(c) 2.5
Ans. (c)
(b) 4/7
(d) 5.2
22r
k
Solution:
Â
r=0
Example 17: The coefficient of x20 in the expansion of
=
7k
( kC r) =
1
k
 Â
k =1 r = 0
320
(c)
20!
Ans. (c)
(d)
60
•
=
Â
k =1
C20
Solution: Note that, coefficient of x20 in
1
1 2
1 20 ˆ 3
Ê
1
+
x
+
x
+
.
.
.
+
x ¯
Ë 1!
2!
20!
= coefficient of x20 in
20
= coefficient of x20 in e3x =
3
20!
7k
22r
7
k
 ( k Cr 4r )
r=0
k
( kC r)
k
57
Ê 5ˆ
Ë 7 ¯ = 1 - 5 7 = 2.5
Example 20: If the 6th term from the beginning is
equal to the 6th term from the end in the expansion of
1 ˆn
Ê 15
2
+
˜ , then n is equal to:
ÁË
31 5 ¯
(a) 7
(c) 10
Ans. (c)
3
1
1 2
1 20
1 21
Ê
ˆ
1
+
+
+
+
+
+
x
x
.
.
.
x
x
.
.
.
Ë 1!
¯
2!
20!
21!
k
1
Ê 5ˆ
(1 + 4)k = Ë ¯
k
7
7
•
Thus,
3
(b)
20!
(a) 0
7.7
(b) 9
(d) 12
Solution: 6th term from the beginning
5
Ê 2n 5 ˆ
Ê 1 ˆ
= nC5 (21/5)n – 5 Á 1 5 ˜ = nC5 ÁË
˜
Ë3 ¯
6 ¯
Also, 6th term from the end
= (n + 2 – 6) = (n – 4) th from the beginning
7.8
Complete Mathematics—JEE Main
n-5
Ê 1 ˆ
Ê 6 ˆ
= nCn–5 (21/5)n – (n – 5) Á 1 5 ˜
= nC 5 Á n 5 ˜
Ë3 ¯
Ë3 ¯
We are given
n
Ê 2n 5 ˆ n
C5 ÁË
˜ = C5
6 ¯
Ê 6 ˆ
n/5
2
ÁË n 5 ˜¯ fi 6 = 6 fi n = 10
3
Solution: Tr + 1, the (r + 1)th term in the expansion of
xˆn
Ê
ÁË 2 + ˜¯ is given by
3
r
Ê 2 n -r ˆ
Ê xˆ
Tr + 1 = nCr 2n – r Á ˜ = nCr Á r ˜ x r
Ë 3¯
Ë 3 ¯
n!
8! (n - 8)!
1
=
fi
7! (n - 7)!
n!
6
fi
48 = n – 7 fi
fi
n
n
C7
=
C8
2 n -8
38
.
37
2n-7
8
1
=
n-7
6
n = 55
1 ˆ 15
Ê
Example 22: In the expansion of Á x 3 - 2 ˜ , the conË
x ¯
stant term equals
(b) 0
(a) 15C9
15
(d) 15C11
(c) – ( C9)
Ans. (c)
Solution: Tr + 1, the (r + 1)th term in the expansion of
15
Ê 3 1ˆ
x
ÁË
˜ is
x2 ¯
n = 6 and 6C3 a 3 = 5/2
5 3! 3! 5 1 1
= ¥
=
a3 = ¥
2
6!
2 20 8
a = 1/2.
fi
\
fi
Thus,
n = 6, a = 1/2.
Example 24: If the 6th term in the expansion of
8
Ê 1
ˆ
2
ÁË 8 / 3 + x log10 x˜¯ is 5600, then value of x is
x
(a) 2
(b) 5
(c) 10
(d) 10
Ans. (d)
Solution: Note that for log 10x to be defined, x > 0.
We have
fi
Ê 1 ˆ
T 6 = T 5 + 1 = 8C 5 Á 8 / 3 ˜
Ëx ¯
5600 =
(x2 log 10x)5
8! Ê 1 ˆ 10
5
Á ˜ x (log 10 x )
5! 3! Ë x8 ¯
y
y = 100/x2
y = 100/x2
y = (log10x)5
To obtain constant term, we set 45 – 5r = 0 fi r = 9
\ Coefficient of constant term is 15C9 (–1)9 = – 15C9.
Example 23: If a is real and the 4th term in the expan1ˆ n
Ê
sion of Á ax + ˜ is 5/2, for each x Œ R – {0}, then values
Ë
x¯
of n and a are respectively
(a) 5, 1/2
(b) 6, – 1/2
(c) 3, 1/3
(d) 6, 1/2
Ans. (d)
8-5
fi
5600 = 56 x2 (log 10x)5
fi
100 = x2 (log 10x)5
fi
100/x2 = (log 10x)5
The curves
y = 100/x2,
y = (log 10x)5
intersect in just one point.
See Fig. 7.1. This points is (10, 1).
Therefore,
x = 10.
r
Ê 1ˆ
Tr + 1 = 15Cr (x3)15 – r Á - 2 ˜ = 15Cr (–1)r x45 – 5r
Ë x ¯
Solution: We have
3
n – 6 = 0 and nC3 a n – 3 = 5/2
xˆn
Ê
Example 21: If the coefficients of x7 and x8 in Á 2 + ˜
Ë
3¯
are equal, then value of n is
(a) 56
(b) 55
(c) 47
(d) 19
Ans. (b)
According to the given condition
Ê 2n-7 ˆ
Ê 2 n -8 ˆ
n
C7 Á 7 ˜ = n C8 Á 8 ˜ fi
Ë 3 ¯
Ë 3 ¯
Ê 1ˆ
T4 = T3 + 1 = nC3 (ax)n – 3 Á ˜
Ë x¯
n
n–3 n–6
= C3 a
x
= 5/2
As this is true for each x ΠR Р{0}, we get
O
1 10
Fig. 7.1
Example 25: If the (r + 1)th term in the expansion of
1/ 3
Êa
b1 / 2 ˆ
+
ÁË b1 / 6 a1 / 6 ˜¯
value of r is
21
has equal exponents of both a and b, then
Mathematical Induction and Binomial Theorem
(a) 8
(c) 10
Ans. (b)
(b) 9
(d) 11
Solution: We have
x + 1 = (x1/3)3 + 1 = (x1/3 + 1) (x2/3 – x1/3 + 1),
x–1=
Solution: We have
Ê a1 / 3 ˆ
Tr + 1 = Cr Á 1 / 6 ˜
Ëb ¯
21- r
21
= 21Cr
a
7-r / 3
b
Ê b1 / 2 ˆ
ÁË a1 / 6 ˜¯
and
r
Thus,
r /2
= 21Cr a7 – r/2 b2r/3 – 7/2
b
ar / 6
Since exponents of a and b in the (r + 1)th term are equal,
7-
7 / 2 -r / 6
.
r
2r 7
21 7
- fi
= r fi r = 9.
=
3 2
2 6
2
Example 26: If A and B are coefficients of xn in the
expansions of (1 + x)2n and (1 + x)2n – 1 respectively, then
(a) A = B
(b) A = 2B
(c) 2A = B
(d) none of these
Ans. (b)
Solution: We know that coefficient of xr in the expansion of (1 + x)m is mCr.
Thus,
A = 2nCn and B = 2n – 1Cn.
We have
A
=
B
fi
2n
Cn
2 n -1
Cn
=
2n
(2 n)! (n!) (n - 1)!
=
=2
n
n! n! (2 n - 1)!
A = 2B
Example 27: If x2k occurs in the expansion of
1 ˆ n -3
Ê
ÁË x + 2 ˜¯ , then
x
(a) n – 2k is a multiple of 2
(b) n – 2k is a multiple of 3
(c) k = 0
(d) none of these
Ans. (b)
Solution: Tr+1 the (r + 1)th term in the expansion of
1 ˆ n -3
Ê
x
+
is given by
ÁË
˜
x2 ¯
r
Ê 1ˆ
Tr+1 = n–3Cr (x)n–3– r Á 2 ˜ = n–3Cr xn–3–3r
Ëx ¯
n -3
1ˆ
Ê
As x2k occurs in the expansion of Á x + 2 ˜ , we must
Ë
x ¯
have n – 3 – 3r = 2k for some non-negative integer r.
fi 3(1 + r) = n – 2k
fi
n – 2k is a multiple of 3.
Example 28: Coefficient of the term independent of x in
x +1
x - 1 ˆ 10
Ê
is
the expansion of Á 2 / 3
Ë x - x1 / 3 + 1 x - x1 / 2 ˜¯
(a) 210
(c) 70
Ans. (a)
(b) 105
(d) 35
7.9
x – x1/2 =
(
)(
x -1
x ( x - 1) ,
x +1
2/3
)
x +1
1/ 3
-
x -1
x - x + 1 x - x1 / 2
= x1/3 + 1 – (1 + x–1/2) = x1/3 – x–1/2
Now, the (r + 1)th term in the expansion of (x1/3 – x–1/2)10 is
Tr+1 = 10Cr (x1/3)10 – r (x–1/2)r (– 1)r
= 10Cr x10/3 – 5r/6 (– 1)r
For this term to be independent of x,
10 5r
10 6
=0 fi r=
¥ =4
3
6
3 5
Thus, coefficient of the term independent of x in the given
expression is
10!
10
= 210.
C4 =
4! 6!
Example 29: If (1 + x)n = C0 + C1x + C2x2 + … + Cnxn,
then value of C0 + 2C1 + 3C2 + … + (n +1)Cn is
(a) 2n –1
(b) n(2n – 1)
n –1
n
(d) (n + 1)2n
(c) n(2 ) + 2
Ans. (c)
Solution: Let E = C0 + 2C1 + 3C2 + … + nCn – 1 +
(1)
(n + 1)Cn
Using Cr = Cn – r , we can rewrite (1) as
E = (n + 1) C0 + nC1 + (n – 1)C2 + … +
2Cn –1 + Cn
(2)
Adding (1) and (2), we get
2E = (n + 2)C0 + (n + 2)C1 + (n + 2)C2 + …
+ (n + 2)Cn
= (n + 2) {C0 + C1 +… + Cn} = (n + 2)2n
fi
E = (n + 2)2n –1
Alternative Solution
We have
C0 x + C1 x2 + C2 x3 + … + Cn xn+1 = x (1 + x)n
Differentiating both the sides, we get
C0 + 2C1 x + 3C2 x2 +… + (n + 1)Cn xn
= (1 + x)n + nx(1 + x)n –1
(1)
Putting x = 1, we get
C0 + 2C1 + 3C2 + … + (n + 1)Cn
= 2n + n(1) 2n –1 = (n + 2)2n – 1
Example 30: If n > 1, and (1 + x)n = C0 + C1 x + C2 x2 +
… + Cn xn then value of C0 – 2C1 + 3C2 – 4C3 + … + (– 1)n
(n + 1)Cn is
Complete Mathematics—JEE Main
7.10
(a) – 1
(c) 1
(b) 0
(d) 2
Ans. (b)
Solution: Putting x = –1 in the equation (1) of alternative solution to Example 29, we get
C0 – 2C1 + 3C2 – 4C3 +… + (–1)n (n + 1)Cn = 0 + 0 = 0
Example 33: The positive integer just greater than S =
(1 + 0.0001)10000 is
(a) 2
(b) 3
(c) 4
(d) 5
Ans. (b)
Solution: Let n = 10000, then
1 n
S =Ê 1 + ˆ
Ë
n¯
Example 31: If n ΠN, n > 1, then value of
E = a – nC1 (a – 1) + nC2 (a – 2) +… + (– 1)n (a – n) (nCn) is
(a) a
(b) 0
2
(d) 2n
(c) a
2
Ê 1ˆ
Ê 1ˆ
Ê 1ˆ
= 1 + nC1 Á ˜ + nC2 Á ˜ + + nCn Á ˜
Ë n¯
Ë n¯
Ë n¯
Ans. (b)
1 n ( n - 1) 1 n ( n - 1) ( n - 2 )
+
2! n2
3!
n3
1 n ( n - 1) (n - n + 1)
++
n!
nn
= 1+1+
Solution: We can write E as
a [nC0 – nC1 + nC2 – … + (–1)n (nCn)] + [nC1 –
(2) (nC2) + (3) (nC3) – …– (– 1)n (n) (nCn)]
= 0 + F where
n
n
1Ê
1ˆ 1 Ê
1ˆ Ê
2ˆ
1 - ¯ + Ë1 - ¯ Ë1 - ¯
Ë
2!
3!
n
n
n
1
1
n - 1ˆ
+ + Ê1 - ˆ Ê1 n! Ë
n¯ Ë
n ¯
= 1+1+
n
F = C1 – (2) ( C2) + (3) ( C3) – … –
(– 1)n (n) (nCn)
We have (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + … + nCn xn
Differentiating, we get
< 1+1+
n(1 + x)n–1 = nC1 + 2(nC2)x + 3(nC3)x2 + … + n(nCn)xn–1
Putting x = – 1, we get
0 = nC1 – 2(nC2) + 3(nC3)– … – (–1)n–1 (n)(nCn)
But
fi
0 = F.
Thus,
Thus,
E = 0 + 0 = 0.
n
 n Cr an – r br cos [nA – (n – r) B]
r=0
is:
(a) (a + b)n
(c) cn
Ans. (c)
(b) (a – b)n
(d) c2n
n
Solution: S =
 n Cr
an – r br Re[ei(nB – (n – r)A]
r=0
n
= Re
 n Cr (ae – iB)n – r (be iA)r
r=0
= Re (ae – iB + beiA)n
= Re [a(cos B – i sin B) + b (cos A + i sin A)]n
= Re [(a cos B + b cos A)
– i (a sin B – b sin A)]n
But a cos B + b cos A = c
[projection formula]
and a sin B = b sin A
[Law of sines]
\
S = Re(cn) = cn.
1 1
1
+ ++
2! 3!
n!
1
1
1
=
£ r -1 " r ≥ 2 .
r!
1.2 … r 2
Example 32: Suppose ABC is a triangle and n is a natural number, then sum of the series
S=
n
S £ 1+1+
= 1+
1 1
1
+ 2 ++ n –1
2 2
2
n -1
1 – (1 / 2 )n
1
=3–Ê ˆ
< 3.
Ë2 ¯
1 – 1/ 2
Example 34: The coefficient of the middle term in the
binomial expansion of (1 + a x)4 and of (1 – ax) 6 is the same
if a equals
3
10
(b)
3
10
5
3
(c) –
(d)
3
5
Ans. (a)
Solution: Middle term in the expansion of (1 + a x)4 is
4
C2(ax)2 = 6a 2x 2 and the middle term in the expansion of
(1 – ax) 6 is 6C3(– a x)3 = – 20 a 3 x3.
(a) –
We are given 6a2 = – 20 a 3 fi a = 0 or a =
Example 35: For natural numbers m and n
(1 – y) m (1 + y)n = 1 + a1 y + a2 y2 + º
and a1 = a2 = 10, then (m, n) equals
(a) (35, 45)
(b) (20, 45)
(c) (35, 20)
(d) (45, 35)
Ans. (a)
Mathematical Induction and Binomial Theorem
Solution: (1 – y)m (1 + y)n = (1 – my + mC2 y 2 – º) (1 + ny
n
+ C2 y + º)
n
m
= 1 + (n – m) y + ( C2 + C2 – mn) y 2 + º
We are given
n – m = a1 = 10
n
fi
fi
fi
\
20
m
C2 + C2 – mn = a2 = 10.
n = m + 10 and
1
1
n (n – 1) + m (m – 1) – mn = 10
2
2
(m + 10) (m + 9) + m(m – 1) – 2m(m + 10) = 20
m2 + 19m + 90 + m2 – m – 2m2 – 20m = 20
– 2m = – 70 fi m = 35.
n = 45
and
fi
Example 36: The sum of the series S =
C2 – 20C3 + º + 20C10 is
(a) – (20C10)
(b)
20
C0 –
20
C1 +
1 20
( C10)
2
(c) 0
(d) 20C10
Ans. (b)
Solution: Using n Cr = nCn – r , we can write
2S = 20C0 – 20C1 + 20C2 – 20C3 + º 20C10 + 20C20 – 20C19 +
20
C18 – º + 20C10
fi
1 20
1
(
C10 ) = ( 20 C10 )
2
2
Example 37: The expansion of
S= 0+
5
(x +
5
) (
)
x 3 – 1 + x – x 3 – 1 is a polynomial of degree
(a) 5
(b) 6
(c) 7
(d) 8
Ans. (c)
Solution: Using (a + b)5 + (a – b)5
= 2[a 5 + 5C2 a3b2 + 5C4 ab4],
(
we get x + x 3 – 1
5
) + (x –
x3 – 1
È
2
5
)
= 2 [x 5 + 10x3(x 3 – 1) + 5x(x3 – 1)2]
which is a polynomial of degree 7.
Example 38: The coefficient of x r in the expansion of
S = (x + 3)n – 1 + (x + 3)n – 2 (x + 2) + (x + 3)n – 3 (x + 2)2+
º + (x + 2)n – 1
is
(a) 3 n – r – 2 n – r
(b) nCr (3 r – 2 r)
n
n–r
n–r
–2 )
(d) none of these
(c) C r (3
Ans. (c)
Solution: We can write
S=
n
7.11
˘
( x + 3)n – 1 Í1 – ÊÁ x + 2 ˆ˜ ˙
Ë x + 3¯
Î
x+2
1–
x+3
˚ = (3 + x) n – (2 + x) n
\ Coefficient of x r = nC r (3n – r – 2n – r )
Example 39: Number of zeros at then end of 991001 + 1 is
(a) 2
(b) 4
(c) 1002
(d) 1004
Ans. (a)
Solution: Let n = 1001. We can write
991001 + 1 = (100 – 1) n + 1
= nC0100 n – nC1 100n – 1 + n C2 100 n – 2 – º
= 100 m + nCn – 1 (100) – 1 + 1
where
m = nC0100n – 1 – nC1 100 n – 2 + n C2 100n – 3 –
n
º – C n – 2 100 + n C n – 1
As n Cn – 1 = 1001, the units digit of m is different from 0.
Thus, the number of zeros at the end of 991001 + 1 is two.
Example 40: The coefficient of xk in the expansion of
E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n is
(a) nCk
(b) n +1Ck
(c) n +1Ck +1
Ans. (c)
(d) none of these
Solution: We have
E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n
=
1 - (1 + x )n+1 1
= ÈÎ(1 + x )n+1 - 1˘˚
1 - (1 + x )
x
Thus, coefficient of xk in E
= Coefficient of xk+1 in [(1 + x)n + 1 – 1]
= n +1Ck + 1
Example 41: The number of irrational terms in the expansion of (51/6 + 21/8)100 is
(a) 96
(b) 97
(c) 98
(d) 99
Ans. (b)
Solution: Tr + 1, the (r +1)th in the expansion of
(51/6 + 21/8)100 is given by
Tr +1 = 100Cr (51/6)100 – r (21/8)r
As 5 and 2 are relatively prime, T r +1 will be rational if
100 - r
r
and are both integers. i.e. if 100 – r is a multi6
8
ple of 6 and r is a multiple of 8. As 0 £ r £ 100, multiples
of 8 upto 100 and corresponding value of 100 – r are
r = 0, 8, 16, 24, …, 88, 96
100 – r = 100, 92, 84, 76, …, 12, 4
Complete Mathematics—JEE Main
7.12
Out of these values for 100 – r, multiples of 6 are 84, 60,
36, 12
\
there are just four rational terms in the expansion of
(5
16
fi
100
+ 21 8 ) .
number of irrational terms is 101 – 4 = 97
Example 42: The sum of the rational terms of
20
(21 / 5 + 3 )
is
(a) 71
(c) 97
Ans. (d)
(b) 85
(d) none of these
Solution: Tr + 1 , the (r + 1)th term in the expansion of
(2
1/ 5
+ 3
20
)
is
Tr +1 = 20Cr (21/5)20 – r (31/2 ) r = 20Cr 24–r/5 (3)r/2
As 2 and 3 are relatively prime 24 – r/5 and 3r/2 will be
rational if both 4 – r/5 and r/2 are integers, i.e. if r/5 and r/2
are integers. This is possible if
r = 0, 10, 20.
Thus, the sum of rational terms is
20
C0(24) (30) + 20C10 (22) (35) + 20C20 (20) (310)
As this is more than 71, 85 or 97, answer is (d).
1 ˆn
Ê
Example 43: If in the expansion of Á x 3 - 2 ˜ , n Œ N,
Ë
x ¯
5
10
sum of the coefficient of x and x is 0, then value of n is
(a) 5
(b) 10
(c) 15
(d) 20
Ans. (c)
Solution: T r + 1, the (r + 1)th term in the expansion of
n
Ê 3 1ˆ
ÁË x - 2 ˜¯ is
x
r
Ê 1ˆ
T r + 1 = nCr(x3)n – r Á - 2 ˜ = nCr (– 1)r x 3n–5r
Ë x ¯
3n - 5
For the coefficient of x5, we set 3n – 5r = 5 fi r =
5
= p(say) and
for the coefficient of x10, we set 3n – 5r = 10 fi r =
3n - 10
= q (say)
5
Note that p – q = 1. We are given
n
Cp(–1) p + nCq (–1)q = 0
n
fi
Cp(–1) p + nCp –1 (–1) p – 1 = 0
n
fi
Cp = nCp –1 fi n – p = p – 1
fi
Thus,
fi
n = 2p–1
fi
n +1
3n - 5
=
2
5
15 = n
p=
fi
n +1
2
5n + 5 = 6n – 10
Example 44: In the expansion of (1 + px)n, n ΠN, the
coefficient of x and x2 are 8 and 24 respectively, then
(a) n = 3, p = 2
(b) n = 4, p = 2
(c) n = 4, p = 3
(d) n = 5, p = 3
Ans. (b)
Solution: We have
(1 + px)n = 1 + nC1(px) + nC2(px)2 + …
1
n(n – 1)p2x2 + …
= 1 + npx +
2
According to the hypothesis,
1
n(n – 1)p2 = 24
np = 8 and
2
Putting p = 8/n in the second expression, we get
2
1
n - 1 24 ¥ 2 3
Ê 8ˆ
n(n – 1) Á ˜ = 24 fi
=
=
Ë
¯
n
2
n
8¥8 4
fi 4n – 4 = 3n fi n = 4
Putting this value in np = 8, we get p = 2.
Example 45: The greatest value of the term independent of x, as a varies over R, in the expansion of
sin a ˆ 20
Ê
x
cos
a
+
ÁË
˜ is
x ¯
(b) 20C15
(a) 20C10
(c) 20C19
(d)
10
Ê 1ˆ
C10 Á ˜
Ë 2¯
20
Ans. (d)
Solution: Tr +1, the (r+1)th term in the expansion of
sin a ˆ
Ê
ÁË x cos a +
˜
x ¯
20
20
is Cr (x cos a)
20–r
Ê sina ˆ
ÁË
˜
x ¯
r
= 20Cr x20–2r (cos a)20 –r (sin a)r
For this term to be independent of x, we set 20 – 2r = 0
fi r = 10.
Let
b = Term independent of x, then
b = 20C10 (cos a)10 (sin a)10
= 20C10 [ cos a sin a]10
10
Ê 1ˆ
Thus, the greatest possible value of b is 20C10 Á ˜ .
Ë 2¯
Example 46: Sum of the last 20 coefficients in the expansion of (1 + x)39, when expanded in ascending powers
of x, is
(b) 218
(a) 219
40
19
(c) C20 – 2
(d) 238
Ans. (d)
Solution: There are 40 terms in the expansion of
(1 + x)39. Sum of the last 20 coefficients is
S = 39C20 + 39C21 + … + 39C38 + 39C39
Mathematical Induction and Binomial Theorem
S = 39C19 + 39C18 + … + 39C1 + 39C0
[using nCr = nCn – r]
Adding the above two expressions, we get
2S = 39C0 + 39C1 + … + 39C39 = 239
fi
S = 238
fi
Example 47: If (1 + x)15 = C0 + C1 x + C2 x2 + … +
C15 x15 then value of the expression S = C2 + 2C3 + 3C4 +
… + 14C15 is
(b) 13(215)
(a) 13(214) + 1
(c) 13(214)
Ans. (a)
Solution: We add –1C0 + 0C1 to both the sides of S and let
Note that
S1 = – 1C0 + 0C1 + 1C2 + 2C3 + … + 13C14 + 14C15
(1)
Using Cr = Cn – r , we rewrite the above expression as
S1 = 14C0 + 13C1 + 12C2 + … + 0C14 + (–1)C15
(2)
Adding (1) and (2), we get
S = 13(214) + 1
fi
Note that
R = [( 2 + 1)6]
2 – 1 < 1 fi 0 < F = ( 2 –1)6 < 1.
We have, R + f + F =
Example 48: Let R = (2 + 3 )2 n and f = R – [R] where
[ ] denotes the greatest integer function, then R(1– f ) is
equal to
(a) 1
(b) 22n
2n
(c) 2 – 1
(d) 2nCn
Ans. (a)
6
)
2 -1
(a) (nCm)3
(b) 3(nCm)
(c) nC3m
(d)
3n
Cm
Ans. (d)
Solution: We have
(1 + x)n (1 + y)n (1 + z)n
Ê n
ˆÊ n
ˆÊ n
ˆ
= Á Â n Cr x r ˜ Á Â n Cs y s ˜ Á Â n Ct z t ˜
Ë r =0
¯ Ë s=0
¯ Ë t =0
¯
<1
=
Â
(n Cr )(n Cs )(n Ct ) x r y s z t
0£ r ,s,t £ n
We have
R + F = (2 +
2n
3 ) + (2 –
2n
2n
2n
= 2 [ C0 2 + C2 2
3)
2n–2
2n
For sum of the coefficients of degree m, we must have
2
( 3)
2n
C4 (22n–4) ( 3 )4 … + 2nC2n
R + F is an even integer.
[R] + f + F is an even integer.
f + F is an integer.
But 0 £ f < 1 and 0 < F < 1
fi
0<f+F<2
But the only integer between 0 and 2 is 1. Thus,
fi
f+F=1
Now,
6
) (
2 +1 +
= 2 [23 + 15(22) + 15(2) + 1] = 198
fi
f + F = 198 – R is an integer.
As 0 £ f < 1 and 0 < F < 1, 0 < f + F < 2.
But the only integer between 0 and 2 is 1. Therefore
f+F=1
Thus,
1 = 198 – R
fi
R = 197
Solution: As 0 < 2 – 3 < 1 we get 0 < F = (2 – 3 )2n
fi
fi
fi
(
Example 50: Sum of the coefficients of the terms of
degree m in the expansion of (1 + x)n (1 + y)n (1 + z)n is
2S1 = 13 [C0 + C1 + C2 + … + C15] = 13(215)
S1 = 13(214)
then [( 2 + 1)6] is equal
(a) 199
(b) 198
(c) 197
(d) 196
Ans. (c)
Solution: Let ( 2 + 1)6 = R + f where R is an integer
and 0 £ f < 1.
= 2 [ 6C 0 ( 2 ) 6 + 6C 2 ( 2 ) 4 + 6C 4 ( 2 ) 2 + 6C 6]
S1 = S – 1C0 + 0C1 = S – 1.
fi
Example 49: If [ ] denotes the greatest integer function,
Also 0 <
(d) none of these
7.13
(
R(1 – f) = RF = 2 + 3
1–f=F
2n
2n
) (2 - 3 )
= (4 – 3)2n = 12n = 1
r + s + t = m where r, s, t are integers with r, s, t ≥ 0.
+
2n
( 3)
Sum of such coefficients
]
=
 ( n Cr ) ( n Cs ) ( n Ct )
r , s ,t ≥ 0
r + s+t = m
=the number of ways of choosing a total
number of m balls out of n distinct black, n
distinct white and n distinct green balls
= 3nCm
Example 51: The remainder when 22000 is divided by
17 is
(a) 1
(b) 2
(c) 8
(d) 12
Ans. (a)
Complete Mathematics—JEE Main
7.14
Solution: We have 24 = 16 = 17– 1
\
2
2000
4 500
= (2 )
=
= (17 – 1)
500
500
C0 17500 – 500C1 17499 + 500C2 17498 –
500
C317497 + … – 500C499 (17) + (–1)500
= 17m + 1
where m is some positive integer.
Thus, the required remainder is 1.
Example 52 : If (1 + x + x2)48 = a0 + a1 x +
a2 x2 + …+ a96 x96, then value of a0 – a2 + a4 – a6 + …
+ a96 is
(a) –1
(b) 0
(c) 1
(d) 48
Ans. (c)
Solution: Putting x = i, we get
(1 + i + i2)48 = a0 + a1 i + a2 i2 + … + a98 i96
fi
i48 = (a0 – a2 + a4 – …) + i(a1 – a3 + …)
Equating the real parts we get
a0 – a2 + a4 – a6 + … + a96 = 1
Example 53: Sum of the series S = 3–1 (10C0) – 10C1 +
(3) (10C2) – 32(10C3) + …+ 39 (10C10) is
(b) 210 – 1
(a) 29
1 11
1 10
(c)
(2 – 2)
(d)
(2 )
3
3
Ans. (d)
Solution: We have
–1 10
10
10
2
10
(a) 2016
(c) 4033
Ans. (b)
Solution: For 1 £ x £ 1013,
x2017 + (2017 – x)2017
= x2017 + 2017C0 (2017)2017
+ 2017C1 (2017)2016 (–x)
+ 2017C2 (2017)2015 (–x)2
+ . . . + 2017C2016 (2017) (–x)2016
+ (–x)2017
= (2017)m
where
m = 2017C0 (2017)2016 + 2017C1 (2017)2015 (–x)
+ . . . + 2017C2016 (–x)2016,
is a positive integer.
Now, write E as sum of 1013 groups as follows:
E = (12017 + 20162017) + (22017 + 20152017)
+ . . . + (10132017 + 10142017)
and note that each group is divisible by 2017.
200
Example 56: Let (1 + x + x2)100 =
and a =
Example 54: The coefficient of x53 in
200
 ar , then value of
Â
(a) 5
(c) 3
Ans. (b)
(1)
r =1
rar
is
25a
(b) 4
(d) 2
Solution: a =
200
200
 ar =
 ar (1) r
r =0
= 3100
r =0
Differentiating (1), we get
100 (1 + x + x2)99 (1 + 2x) =
200
 rar
xr – 1
r =1
Put
x = 1, to obtain
200
100
100(3)99 (3)=
 100 Cm ( x - 5)100- m (4)m is
m =0
(a) 100C47
(c) – 100C53
Ans. (c)
(b) 100C53
(d) 253
100
 100 Cm ( x - 5)100- m (4)m
 rar
r =1
Solution: We have
= (x – 5 + 4)100 = (x – 1)100
m =0
\
200
r =0
S = 3 [ C0 – C1 (3) + C2 (3 ) – C3 (3 )
1
1
(1– 3)10 = (210)
=
3
3
 ar xr
r =0
3
+… + 10C10(310)]
(b) 2017
(d) 20162
Coefficient of x53 is 100C53 (–1)47 = – 100C53.
Example 55: Let
E = 12017 + 22017 + 32017 + . . . + 20162017,
then E is divisible by
200
fi
Â
r =1
rar
100a
=
=4
25a
25a
Example 57: If ai (i = 0, 1, 2, . . . 16) are real constants
such that for every real value of x
(1 + x + x2)8 = a0 + a1x + a2x2 + . .. + a16x16, then a5 is
equal to
(a) 502
(b) 504
(c) 506
(d) 508
Ans. (b)
r
8!
 p ! q ! r ! 1p x p ( x 2 )
Solution: (1 + x + x2)8 =
p, q , r ≥ 0
P +q +r =8
Mathematical Induction and Binomial Theorem
For a5, we require, q + 2r = 5,
fi
q = 5, r = 0, p = 3 or q = 3, r = 1, p = 4,
or
q = 1, r = 2, p = 5.
Thus, coefficient of x5 is
8!
8!
8!
+
+
= 504
5! 3! 3! 1! 4! 1! 2! 5!
a
c
Ê b ˆ
+
- 2Á
is equal to
Ë b + c ˜¯
a+b c+d
(a) n
(b) 1
(c) 0
(d) 2n
Ans. (c)
Example 58: Sum of the last 12 coefficients of in the
binomial expansion of (1 + x)23 is:
(b) 223
(a) 222
23 23
(c) 2 – C11
(d) 223 – 211
Ans. (a)
Solution: Sum of last 12 coefficients is
S = 23C12 + 23C13 + . . . + 23C23
n
Cr = nCn – r, we get
Using
S = 23C11 + 23C10 + . . . + 23C0
Adding (1) and (2), we get
2S = 23C0 + 23C1 + . . . 23C23 = 223
fi
S = 222
3
Solution: Suppose a = nCr, b = nCr + 1, C = nCr + 2, and
d = nCr + 3. Now
a
=
a+b
Similarly,
304 19
- a = 0 fi a = 16
3
3
544 272
=
3
3
Example 60: Let a, b, c, d be any four consecutive coefficients in the binomial expansion of (1 + x)n, then
n
Cr + Cr + 1
n
=
n +1
Cr
Cr + 1
=
r +1
n +1
Example 61: Suppose a0 = 2017, a1 a2, . . . , an – 1, 2023
= an are in A.P. Let
S=
4
Solution: S = (1 + ax + bx2) (1 – 2x)18
= (1 + ax + bx2) [1 + 18C1 (–2x) +
18
C2 (–2x)2 + 18C3 (–2x)3 + 18C4 (–2x)4 + . . .]
3
Coefficient of x in the expansion of S is 18C3 (– 2)3 +
a(18C2 (–2)2) + 18C1(–2)b = 0
Divide by 18C1(–2) to obtain
544
– 17a + b = 0
(1)
3
Similarly, coefficient of x4 is
18
C4 (–2)4 + a (18C3 (–2)3) + 18C2 (–2)2 b = 0
18
Divide by C2 (–2)2 to obtain
32
80 –
a+b=0
(2)
3
Subtract (2) from (1) to obtain
n
Cr
c
a
Ê b ˆ
- 2Á
+
=0
Ë b + c ˜¯
c+d
a+b
Thus,
(2)
n
(r + 1) + 1
(r + 2 ) + 1
b
c
=
and
=
b+c
n +1
c+d
n +1
(1)
Example 59: If coefficient of x and x in the expansion
of (1 + ax + bx2) (1 – 2x)18 in powers of x are both zeros,
then (a, b) is equal to
251ˆ
272 ˆ
Ê
Ê
(b) Ë 14,
(a) Ë 14,
¯
3
3 ¯
272 ˆ
251ˆ
Ê
Ê
(c) Ë 16,
(d) Ë 16,
¯
3
3 ¯
Ans. (c)
From (1), b = 17 ¥ 16 –
7.15
1
2n + 1
Then S is equal to
(a) 10
(c) 1013
Ans. (a)
n
 ( n Cr )
ar – 1000
r=0
(b) 40
(d) 2021
Solution: Put nCr = Cr and let
T = a0 C0 + a1 C1 + a2 C2 + . . . + an Cn (1)
Using
Cr = Cn – r, we get
(2)
T = an C0 + an – 1 C1 + . . . + a0 Cn
Adding (1) and (2), we get
2T = (a0 + an)C0 + (a1 + an–1)C1 + (a2 + an–2)C2
+ . . . + (an + a0)Cn
But a0 + an = a1 + an – 1 = a2 + an – 2
= . . . = an + a0 = 2017 + 2023
\
2T = 4040 (C0 + C1 + . . . + Cn)
fi
T = (2020)2n = (1010)2n + 1
Thus,
1
S = n + 1 (T ) – 1000 = 10
2
Example 62: The interval in which x (> 0) must
lie so that the greatest term in the expansion of
(1 + x)2n has the greatest coefficient is
Ê n -1 n ˆ
(a) Á
,
Ë n n - 1˜¯
Ê n n + 1ˆ
,
(b) Á
Ë n + 1 n ˜¯
Ê n n + 2ˆ
,
(c) Á
Ë n + 2 n ˜¯
(d) none of these
Ans. (b)
7.16
Complete Mathematics—JEE Main
Solution: The greatest coefficient in the expansion of
(1 + x)2n is 2nCn. We are given 2nCn xn is the greatest term.
2n
Cn –1 xn–1 < 2nCn xn and
2n
Cn +1 xn+1 < 2nCn xn
\
2n
fi
Cn -1
2n
2n
<x<
Cn
2n
Cn
Cn +1
n
n +1
<x<
.
n +1
n
fi
Example 63: If the coefficients of the r th, (r + 1)th and
(r + 2)th terms in the binomial expansion of (1 + y)m are in
A.P., then m and r satisfy the equation
(a) m 2 – m(4r + 1) + 4r 2 – 2 = 0
(b) m 2 – m(4r – 1) + 4r2 + 2 = 0
(c) m 2 – m(4r – 1) + 4r2 – 2 = 0
(d) m2 – m(4r + 1) + 4r2 + 2 = 0.
Ans. (a)
Solution: Coefficient of the r th term is m Cr – 1 .
According to the given condition mCr – 1 + mCr + 1 = 2(mCr)
m
fi
fi
fi
fi
Cr –1
m
Cr
m
+
Cr + 1
m
Cr
r
m–r
+
=2
=2fi
m +1– r r +1
r 2 + r + (m – r) (m + 1 – r) = 2(m + 1 – r) (r + 1)
r2 + r + m2 + (1 – 2r) m – r (1 – r) =
2m (r + 1) + 2(1 – r 2)
m 2 – m(4r + 1) + 4r 2 – 2 = 0.
Example 64: Let x > – 1, then statement
P(n) : (1 + x) n > 1 + nx
is true for
(a) all n ΠN
(b) all n > 1
(c) all n > 1 provided x π 0
(d) none of these.
Ans. (c)
Solution: For n = 2,
P(2): (1 + x)2 = 1 + 2x + x2 > 1 + 2 x as x π 0.
Assume that
P(k): (1 + x) k > 1 + kx
(1)
for some k ΠN, k > 1.
As x > – 1, multiplying both the sides of (1) by 1 + x, we get
(1 + x) k + 1 > (1 + kx) (1 + x)
= 1 + (k + 1)x + k x 2
> 1 + (k + 1)x
[
k x 2 > 0]
Thus, P(k + 1) is true.
By the principle of mathematical induction P(n) is true for
all n > 1 provided x π 0.
Example 65: For each n ΠN, 23n Р1 is divisible by
(a) 8
(b) 16
(c) 32
(d) none of these
Ans. (d)
Solution: We have
23n – 1 = (23)n – 1 = (1 + 7)n – 1
= 1 + nC1 (7) + nC2 (72) + … + nCn (7n) – 1
= 7m
where
m = nC1+ nC2 (7) +…+nCn (7n – 1) is a positive integer.
3n
Thus, 7 divides 2 Р1 for all n ΠN.
Example 66: For each n Œ N, x2n–1 + y2n –1 is divisible by
(a) x + y
(b) (x + y)2
(c) x3 + y3
(d) none of these
Ans. (a)
Solution: x2n–1 + y2n–1
= (x + y)(x2n–2 – x2n –3y + x2n – 4 y2– x2n–5y3
+…+ y2n – 2)
Thus x2n – 1 + y2n –1 is divisible by x + y for all n Œ N.
Example 67: For all n ΠN, n(n + 1) (2n + 1) is divisible by
(a) 6
(b) 8
(c) 15
(d) 9
Ans. (a)
Solution: We have n(n + 1) (2n + 1) = 6{12 + 22 + …+
n2},
\
n(n + 1) (2n + 1) is divisible by 6 for all n ΠN.
Example 68: If an = 7 + 7 + 7 + having n radical signs then by methods of mathematical induction which
of the following is true
(b) an > 3 V n ≥ 1
(a) an > 7 V n ≥ 1
(c) an < 4 V n ≥ 1
(d) an < 3 V n ≥ 1
Ans. (c)
Solution: We have a1 = 7 < 4. Assume ak < 4 for
some natural number k.
We have
a k + 1 = 7 + ak < 7 + 4 < 4
\ an < 4 V n ≥ 1.
n
Example 69: If an = 22 + 1, then for n > 1, last digit
of an is
(a) 5
(b) 7
(c) 3
(d) 4
Ans. (b)
Solution: For n = 2, an = 24 + 1 = 16 + 1 = 17. Assume
k
that ak = 22 + 1 = 10m + 7 where k > 1 and m is some positive integer.
Mathematical Induction and Binomial Theorem
2k + 1
Now,
2k 2
ak + 1 = 2
+ 1 = (2 ) + 1
= (10m + 6)2 + 1 = 10 (10m2 + 12m + 3) + 7
Thus, last digit of an is 7 V n > 1.
Example 70: Let S (K): 1 + 3 + 5 + º + (2K – 1) = 3 +
K . Then which of the following is true?
(a) S (K) fi
/ S (K + 1)
(b) S (K) fi S (K + 1)
(c) S(1) is correct
(d) principle of mathematical induction can be used to
prove the formula.
Ans. (b)
2
Solution: S (1): 1 = 3 + 12 which is not true. Suppose
S(K) is true, then
1 + 3 + 5 + º + (2K – 1) = 3 + K 2
Adding (2K + 1) to both the sides, we get
1 + 3 + 5 + º + (2K – 1) + (2K + 1) = 3 + K2 + 2K + 1
= 3 + (K + 1)2
which is S (K + 1).
Thus, S(K) fi S(K + 1).
Example 71: Let (1 + x) n = C0 + C1 x + C2 x2 + º + Cn xn
C
C
C
C
1
and 1 + 2 2 + 3 3 + + n n = n (n + 1) , then
C0
C1
C2
Cn – 1 k
value of k is
(a) 2
(c) 6
Ans. (a)
(b) 3
(d) 12
Solution: r
n
Âr
fi
r =1
\
Cr
n!
(r - 1)! (n - r + 1)!
= r
r ! ( n - r )!
n!
Cr – 1
=n–r+1
Cr
=
Cr – 1
n
1
1 ˆ
of ÊÁ 2 –
˜
Ë
2¯
beginning is
(a) 10C6
(c)
n
log38
, then the 5th term from the
(b) 2(10C4)
10
4
) + ( x2 +
1 - x3
)
4
, is
(b) 8
(d) 12
Solution: We have
(x – y)4 + (x + y)4 = 2 [4C0 x4 + 4C2 x2 y2 + 4C4 y4]
Thus,
( x2 -
4
1 - x3 ) + ( x2 + 1 - x3 )
4
= 2[4C0(x2)4 + 4C2(x2)2 (1 – x3) + 4C4(1 – x3)2]
which is a polynomial of degree 8.
1
in the expansion of
Example 74: Coefficient of
x
(1 + x)n (1 + 1/x)n is
(a) 2n Cn – 1
(b) 2nCn
(c) 1
(d) 0
Ans. (a)
Solution: We have
1 ˆ n (1 + x )2 n
Ê
(1 + x )n Ë 1 + ¯ =
x
xn
1ˆ
Ê
\ Coefficient of x–1 in (1 + x)n Ë 1 + ¯
x
n
= Coefficient of xn – 1 in (1 + x)2n
= 2nCn–1
Example 75: Coefficient of x4 in (1 + x + x2 + x3)11 is
(a) 99
(b) 330
(c) 990
(d) 1050
Ans. (c)
Solution: We have
fi
1 + x + x2 + x3 = 1 + x + x2(1 + x) = (1 + x)(1 + x2)
(1 + x + x2 + x3)11 = (1 + x)11(1 + x2)11
= (1 + 11 C1 x + 11C2 x2 + 11C3 x3 +
11
C4x4 + ) (1 + 11C1 x2 +
11
C2 x4 + )
4
Thus, coefficient of x in (1 + x + x2 + x3)11
= (1) (11C2) + (11C2) (11C1) + (11C4) (1)
= 55 + (55)(11) + 330 = 990
Example 76: The sixth term in the expansion of (x1/3 +
y ) , if the binomial coefficient of the third term form the
end is 45, is
(b) 45
(a) 252 x5/3 y5/2
5/3 5/2
(c) 45 x y
(d) none of these
Ans. (a)
1/2 n
8
2
1
1 ˆn
Cn ÊÁ –
= (3–1 – 2/3 )log3 = 3– 5 log3 = 5
˜
¯
Ë
2
2
fi
n = 10.
\ 5th term from the beginning
=
1 - x3
(a) 7
(c) 9
Ans. (b)
(d) – 10C6
Ans. (b)
Solution:
( x2 -
r =1
is Ê 1 ˆ
1 ˜
ÁË
(3 ) 9 3 ¯
1 10
( C 4)
2
Example 73: Degree of the polynomial
 (n - r + 1) = 2 n (n + 1)
k = 2.
Example 72: The last term in the binomial expansion
7.17
n
4
Ê 1 ˆ = 2(10C )
˜¯
C4 ( 2 )10 – 4 ÁË –
4
2
Solution: Binomial coefficients of the third term from
the end in the expansion of (x1/3 + y1/2)n, is
7.18
Complete Mathematics—JEE Main
n
C2 = 45 fi n (n – 1) – 90 = 0 fi n = 10.
\ Sixth term in the expansion of (x1/3 + y1/2)n is
n
C5 (x1/3)n–5 (y1/2)5 = 252 x5/3 y5/2
2
3
(a) 128
(c) 32
Ans. (b)
(a) ( 4 + 7 ) / 3
Solution: Coefficient of xr in the expansion of (1 + x/a)6
= 6Cr (1/a)r
According to given condition 6C2(1/a)2, 6C3(1/a)3, 6C4(1/a)4
are in A.P.
3
2
Ê 1ˆ
Ê 1ˆ
= 6 C2 Ë ¯ + 6 C 4 Ë ¯
a
a
4
= 15(a2 + 1)
=0
= ( 4 + 7 ) / 3 as a > 0.
Example 78: Sum of two middle terms in the expansion
of (1 + x )2n–1 is
(b) 2n – 1Cn
(a) 2n – 1Cn – 1
2n
(c) Cn
(d) 2nCn + 1
Ans. (c)
Solution: Two middle terms in the expansion of (1 + x) 2n-1
Ê 2n - 1 + 1 ˆ
Ê 2n - 1 + 1ˆ
+ 1¯ th terms, that is, nth
are Ë
th and Ë
¯
2
2
and (n +1)th terms.
Coefficients of the middle terms in the expansion of
(1 + x) 2n – 1 are 2n–1Cn–1 and 2n–1Cn and their sum = 2n–1Cn–1
+ 2n–1Cn = 2nCn
Example 79: If a is a real number and if the middle term
8
Êa
ˆ
of Ë + 3¯ is 1120, then value of a is
3
(a) ±2
(b) ±1
(c) ± 3
(d) ± 2
Ans. (a)
8
Êa
ˆ
Solution: Middle term of Ë + 3¯ is the
3
5th term. We have
4
Ê aˆ
4
T5 = T4+1 = 8 C4 Ë ¯ (3) = 70a 4
3
We are given T5 = 1120 fi 70a4= 1120
fi
a4= 16 =24 fia = ± 2
Ê8 ˆ
Ë 2 + 1¯ th or
Example 80: Suppose F is the fractional part of M =
(
(
(
(
6
13 - 11 ) ,
2
13 - 11 ) =
Also, M + N =
(d) 2 + 3
Ê 1ˆ
2 ( 6 C3 ) Ë ¯
a
fi
2 (20) a
fi
3a2 _ 8a + 3
fi
a
As
(b) ( 4 + 3 ) / 3
(c) 2 - 3
Ans. (a)
\
Solution: Let N =
4
Example 77: If a > 0 and coefficient of x , x , x in the
expansion of (1 + x/a)6 are in A.P., then a equals
(b) 64
(d) 16
13 + 11
6
13 + 11 ) +
6
(
13 - 11 ) ,
6
4
2
= 2 È 6 C0 ( 13 ) + 6C2 ( 13 ) ( 11 ) +
ÎÍ
2
4
6
6
C4 ( 13 ) ( 11 ) + 6C6 ( 11 ) ˘˙
˚
fi M + N is an integer, say, J
Let M = K + F, where K is the greatest integer contained in M.
We have
J=M+N= K+N+ F
As 0 < N < 1, 0 < F < 1 we get, 0 < N + F < 2
Also, N + F is an integer
fi N+F=1fi1– F=N
Thus,
6
M (1– F) = ( 13 + 11 )
(
6
13 - 11 ) = 26 = 64
Example 81: The remainder when 82n– 622n+1 is
divided by 9 is
(a) 7
(b) 8
(c) 0
(d) 2
Ans. (b)
Solution: 82n – 62 2n+1
= (9– 1)2n–(63–1)2n+1
= (2nC0 92n–2nC192n–1 + 2nC292n–2 –
2n
2n
C2n–1 (9) + C2n ) – (
+ 2n+1C2 (63 2n–1)–
2n+1
C0(63
–
2n+1
) – 2n+1 C1(632n)
– + 2n+1C1(63) – 2n+1C0)
= 9m + 1+1 = 9m + 2 for some integer m
Thus, when 82n – 62 2n+1 is divided by 9, the remainder is 2.
Example 82: The coefficient of x7 in the expansion of
(1 – x – x2 + x3)6 is
(a) 132
(b) 144
(c) – 132
(d) – 144
Ans. (d)
Solution: (1 – x – x2 + x3)6 = (1 – x)6 (1 – x2)6
coefficient of x7 in (1 – x)6 (1 – x2)6
= coefficient of x7 in [1 – 6C1 x + 6C2 x2–
+ 6C 6 x 6]
+ 6C6 x12]
¥ [1 – 6C1 x2 + 6C2 x4 –
= (–6C1) (–6C3) + (–6C3) (6C2) + (–6C5) (–6C1)
= – 144.
Example 83: If n ΠN, then
6
13 + 11 ) , then value of M (1– F) is
, we get 0 < N <1
(a) an odd positive integer
(b) an even positive integer
(
2n
)
3 +1
-
(
3 -1
2n
)
is
Mathematical Induction and Binomial Theorem
(a) a rational number but not a natural number
(d) an irrational number
Ans. (d)
Solution:
(
2n
)
2n
( 3 - 1)
2 n -1
2 n -3
2 ÈÍ 2 n C1 ( 3 )
+ 2 nC3 ( 3 )
Î
=
+ + 2 nC2 n -1 ( 3 )˘˚
2n
n -1
2n
n-2
= 2 3 ÈÎ C1 (3 ) + C3 3
3 +1
-
+ +
2n
C2 n -1 ˘˚
which is an irrational number.
Example 84: The coefficient of xn in the expansion of
1
1 2
1 nˆ2
Ê
ÁË 1 + x + x . . + x ˜¯
n!
1!
2!
2n
(a)
n!
Example 85: The coefficient of xn in the expansion of
(1 + x) (1 – x)n is
(b) (– 1)n (1 – n)
(a) (– 1)n n
n–1
(d) (n – 1)
(c) (– 1) (n – 1)
Ans. (b)
Solution: (1 + x) (1 – x)n = [2 – (1 – x)] (1 – x)n
= 2 (1 – x)n – (1 – x)n + 1
n
\ coefficient of x in (1 + x) (1 – x)n
= 2 (–1)n – n + 1 Cn (–1)n
= ( – 1)n (2 – n – 1) = (–1)n (1 – n)
Example 86: Sum of the coefficients of x3 and x6
1ˆ 9
Ê
in the expansion of Á x 2 - ˜ is
Ë
x¯
Ans. (a)
r
Solution: Tr + 1 = 9Cr (x2)9 – r ÊÁ - 1 ˆ˜ = (–1)r (9Cr)x18 – 3r
Ë
¯
x
For coefficient of x3, x6, we set 18 –3r = 3,6
fi
r = 5,4.
\ sum of coefficients of x3 and x6
= (–1)5 (9C4) + (–1)4 (9C5) = 0
2n
(b)
n
1
(d)
n!
(c) n!
7.19
Ans. (a)
Example 87: For n ≥ 2, let
n
n
Solution: Coefficient of x in
an =
1
1 2
1 nˆ2
Ê
1
+
x
+
x
+
.
.
.
+
x
Ë 1!
2!
n! ¯
= coefficient of xn is
r =0
1
Cr2
n
bn =
of
2
1
1 2
1 n
1
ˆ
Ê
n +1
.
.
.
1
+
+
+
+
+
x
x
x
x
.
.
.
˜¯
ÁË 1!
(n + 1)!
2!
n!
Ê
2 2 x 2 23 x 3 ˆ
= coefficient of xn in e2x = ÁË 1 + 2 x +
+
. . .˜¯
2!
3!
2n
=
n!
(a)
1
n
2
Coefficient of x in
1
1
+
(n - 1)!1! n!
1 n
2n
ÈÎ C0 + nC1 + nC2 + nCn-1 + nCn ˘˚ =
n!
n!
1
2
r Cr2
equals
an
(b)
1
n2
an – 1
(d) an2
Solution: For r ≥ 1, r Cr = r (nCr) = n (n – 1 Cr – 1)
Â
r =1
1
1 2
1 nˆ
Ê
ÁË 1 + x + x + . + x ˜¯
n!
1!
2!
1
1
1 nˆ
Ê
2
ÁË 1 + x + x + + x ˜¯
n!
1!
2!
1
1
1
+
+ . +
= +
n! 1! ( n - 1)! 2! ( n - 2 )!
, then value
(c) an
Ans. (b)
Thus, bn =
n
Â
r =1
n
Alternate Solution
=
Â
1
n
(
2 n -1
Cr
2
)
=
1
n2
an – 1
1 ˆn
Ê
Example 88: If in the expansion of Á x 3 - 2 ˜ the
Ë
x ¯
5
10
sum of the coefficients of x and x is 0, then the coefficient
of x20 is:
(b) –20C6
(a) 20C6
15
(c) C5
(d) –15C5
Ans. (d)
Solution: (r + 1)th term in the expansion of
n
Ê 3 1ˆ
x
˜ is
ÁË
x2 ¯
r
Ê 1ˆ
Tr + 1 = nCr(x3)n–r Á - 2 ˜
Ë x ¯
7.20
Complete Mathematics—JEE Main
= nCr x3n – 5r (–1)r
For coefficient of x10, set 3n – 5r = 5
3
fi
r = n – 1 = r1 (say)
5
For coefficient of x10, set 3n – 5r = 10
3
fi
r = n – 2 = r2(say)
5
Note that r1 = r2 + 1.
We are given
n
Cr1 (–1)r1 + nCr2 (–1)r2 = 0
n
n
Cr2 = Cr2 + 1 [ r1 = r2 + 1]
fi
Solution: a1 = coefficient of x in
(1 + x) (1 + x + x2) (1 + x + x2 + x3) . . . (1 + x + . . . + x n)
= coefficient of x in (1 + x ) (1 + x ) . . . (1 + x )
n times
n
= coefficient of x in (1 + x)
= nC 1 = n
Example 90: If xn = a0 + a1 (1 + x) + a2(1 + x)2 + . . .
+ an (1 + x)n
= b0 + b1 (1 – x) + b2 (1 – x)2 + . . . + bn(1 – x)n
then for n = 201, (a101, b101) is equal to:
(a) (–201C101, –201C101)
(b) (201C101, – 201C101)
(c) (–201C101, 201C101)
(d) (201C101, 201C101)
Ans. (b)
Solution: xn = [(1 + x) – 1]n = [1 – (1 – x)]n
1
(n – 1)
2
3
1
1
1
3
n –2= n–
fi
n=
fi n = 15
fi
5
2
2
10
2
For coefficient of x20, set 3n – 5r = 20
fi
5r = 45 – 20 = 25 or r = 5.
Thus, coefficient of x20 is – 15C5
fi r2 + r2 + 1 = n fi r2 =
2
2
n
=
 n Ck (1 + x)n – k (–1)k
k=0
n
=
 n Ck (–1)k (1 – x)k
k=0
3
Example 89: If (1 + x) (1 + x + x ) (1 + x + x + x ) . . .
(1 + x + x2 + . . . + xn)
= a 0 + a 1x + a 2x 2 + . . . + a m x m,
then value of a1 is
(a) m + 1
(b) n + 1
(c) n
(d) m
Ans. (c)
ak = coefficient of (1 + x)k in
\
n
 n Ck
(1 + x)n – k (–1)k = (– 1)n – k nCn – k = (– 1)n – k nCk
k=0
and
For
bk = nCk (–1)k
n = 201, k = 101, we get
(a101, b101) = (201C101, – 201C101)
Assertion-Reason Type Questions
Example 91:
Statement-1: For each natural number n, (n + 1)7 – n7 – 1
is divisible by 7
Statement-2: For each natural number n, n7– n is divisible
by 7.
Ans. (a)
Solution: For Statement-2, write n = 7r + k, k = – 3, – 2,
– 1, 0, 1, 2, 3
Now,
n7 – n = (7r + k)7 – (7r + k)
Using binomial theorem, we get
n7 – n – (k7 – k) is divisible by 7.
¤
n7 – n is divisible by 7 if and only if k7 – k is divisible by 7
For
k = ± 3, ± 2, ± 1, k7 – k is divisible by 7 if and
only if k6 – 1 is divisible by 7.
6
But
k – 1 = 728, 63, 0 for k = ± 3, ± 2, ± 1.
\
k6– 1 is divisible by 7 for k = ± 3, ± 2, ± 1.
Thus, n7 Рn is divisible by 7 for each n ΠN.
fi
(n + 1)7 – (n + 1) – (n7 – n) is divisible by 7
fi
(n + 1)7 – n7 – 1 is divisible by 7.
Therefore, both the statements are true and statement - 2 is
a correct reason for Statement-1.
10
Example 92: Let S1 =
 j ( j - 1) ( 10 C j )
j =1
10
S2 = Â j
j =1
10
( 10 C j )and S3 = Â ( j 2 ) ( 10 C j )
j =1
Statement-1
S3 = 55 ¥ 29
Statement-2
Ans. (c)
S1 = 90 ¥ 28 and S2 = 10 ¥ 28
Solution: We have
Mathematical Induction and Binomial Theorem
10
 ( 10 C j ) x j
(1 + x)10 =
j =0
10
 j ( 10 C j ) x j–1
(1)
j =1
Again differentiating both the sides with respect to x, we get
10
8
(10) (9) (1 + x) =
˘
1 È 2n + 2
– 1˙
Í
n +1În + 2 ˚
C
C
C
n Cn
=0
Statement-2: 0 – 1 + 2 – + ( –1)
n+2
2
3
4
=
Differentiating both the sides with respect to x, we get
10(1 + x)9 =
 j ( j - 1) (
10
)
Cj x
j–2
(2)
j =1
Ans. (b)
Solution: From (1)
x
Ú0
(1 + t)n dt =
Putting x =1 in (1) and (2) , we get
10
 j(
S2 =
10
)
and
C j = 10 (2 )
=
 j ( j - 1)
S1 =
(
10
)
C j =(10) (9) (28)=(90) (28)
j =1
Adding the above two equation, we get
x
 [ j + j ( j - 1)](10 C j )
S 3=
= (10) (2 ) (2 + 9)
fi
S3 = (55) (29)
Thus, Statement-1 is true but Statement-2 is false.
2
Solution C0 + C1 + º + Cn
= C 0C n + C 1C n – 1 + º + C nC 0
= number of ways of choosing n persons out of
n men and n women
= 2nCn
\ Statement-2 is true
Let
S = 5C02 + 7C12 + 9C22 + º + (5 + 2n)Cn2 (1)
Using Cr = Cn – r , we can rewrite (1) as
S = (5 + 2n) C02 + (3 + 2n) C12 + . . . + 5Cn2
Adding (1) and (2), we get
(2)
2S = (10 + 2n) (C02 + C12 + . . . + Cn2)
(2n)
fi
S = (5 + n) ( Cn) = (5 + n)
n! n!
Example 94: Let
(1 + t)n = C0 + C1t + C2t2 + º + Cn t n
(2)
x
x (1 + x )
n +1
[C0t + C1t 2 + . . . + Cntn + 1] dt
Ú0
–
(1 + x )n + 2
(n + 1) (n + 2 )
C0 2 C1 3 C2 4
C
x +
x +
x + . . . + n x n+2 .
(3)
n+2
2
3
4
Putting x = – 1 in (3), we obtain that the second statement
is true.
Putting x = 1 in (2) and (3) and subtracting, we obtain that
Statement-1 is also true.
=
Example 93: Let (1 + x)n = C0 + C1x + C2 x2 + º + Cnxn.
Statement-1: 5C02 + 7C12 + 9C22 + º + (5 + 2n)Cn2
= (5 + n ) (2n )!
n! n!
2
2
Statement-2: C0 + C1 + º + Cn2 = 2nC n
Ans. (a)
2
t(1 + t)n dt =
fi
8
2
C0
C
C
x + 2 x2 + + n xn + 1
n +1
1
2
n +1
j =1
[C0 + C1t + C2t 2 + º + Cntn] dt
Multiplying (1) by t and integrating, we get
Ú0
10
x
Ú0
1
[(1 + x)n + 1 – 1]
n +1
fi
9
j =1
10
7.21
Example 95: Let (1 + x)n = C0 + C1x + C2x2 + º + Cnxn.
Statement-1: For m ≥ 2, C0 – C1 + C2 – º +
(– 1)m – 1 Cm – 1 = (– 1)m – 1 (n – 1Cm – 1)
Statement-2: nCr = n – 1Cr – 1 + n – 1Cr for 1 £ r £ n – 1
Ans. (a)
Solution: C0 = n – 1C0
– C 1 = – n – 1C 0 – n – 1C 1
C 2 = n – 1C 1 + n – 1C 2
– C 3 = – n – 1C 2 – n – 1C 3
ºººººººº
(– 1)m – 1(Cm – 1) = (– 1)m – 1(n – 1Cm – 2 + n – 1Cm – 1)
Adding the above equations, we get
C0 – C1 + C2 – C3 + º + (– 1)m – 1Cm – 1
= (– 1)m – 1 (n – 1Cm – 1)
Example 96: Let (1 + x)n = C0 + C1x + C2x2 + º + Cnxn.
2n
Statement-1: S = C0 + (C0 + C1) + (C0 + C1 + C2) + º +
(C0 + C1 + º + Cn – 1) = n(2n – 1)
(1)
C0
Cn
C1
C2
+
+
++
Statement-1:
(1)(2) (2)(3) (3)(4)
(n + 1) (n + 2 )
n
Statement-2:
 Â
j =1 i < j
Ans. (c)
(Ci + Cj) = (n + 1)2n.
7.22
Complete Mathematics—JEE Main
Solution: We can write
S = nC0 + (n – 1)C1 + (n – 2)C2 + º + 1Cn – 1 + 0Cn
n
 an- j b j
Statement-2:
j=0
(1)
(2)
Adding (1) and (2) we obtain
2S = n[C0 + C1 + C2 + º + Cn] = n(2n)
fi
S = n(2n – 1)
n
 an- j b j
Solution:
abn – 1 + bn
=
(Ci + Cj) each Ci (0 £ i £ n) oc-
j =1 i < j
curs exactly n times. Thus
n
n
 Â
j =1 i < j
( )
k=0
Example 97: Let (1 + x)3n = C0 + C1x + C2x2 + º +
C3nx3n, and w π 1 be a cube root of unity
Statement-1: C0 + C1w + C2w2 + C3 + C4w + C5w2 +
= (–1)n
Statement-2: Cube roots unity form a triangle of area 3
square units.
Ans. (c)
Solution: C0 + C1w + C2w2 + C3 + C4w + C5w2 + º
[sum of a G.P.]
a n +1 - b n +1
a-b
Also, coefficient of xm in (a x + b )m + 1 is (m + 1)a m b
\ Statement-2 is true.
n
Next,
Ck wk = (1 + w)3n
k=0
 (-1) j
(2x + 3)n – j (5 – 2x)j
j=0
n
=
 ( 2 x + 3 )n - j
(2x – 5)j
j=0
=
3n
Â
a n ÈÎ1 - (b a )n +1 ˘˚
1- b a
=
(Ci + Cj) = n  Ck = n 2 n
=
= an + an – 1 b + an – 2 b2 + . . . +
j=0
n
 Â
a n +1 - b n +1
, a π b,
a-b
and coefficient of xm in (a x + b ) m + 1 is (m + 1)a m b
Ans. (a)
Using Cr = Cn – r , we can rewrite (1) as
S = 0C0 + 1C1 + 2C2 + º + (n – 1)Cn – 1 + nCn
In the expression
=
(2 x + 3)n +1 - (2 x - 5)n +1
( 2 x + 3) - ( 2 x - 5)
1
[(2x + 3)n + 1 – (2x – 5)n + 1]
8
Coefficient of xn is
1 n+1
[ Cn 2n (3) – n + 1Cn 2n (– 5)]
8
= (n + 1) 2n
\ Statement-1 is true and Statement-2 is correct explanation for it.
=
= (– w2)3n = (– 1)3n w6n
= (– 1)n (1) = (– 1)n.
Statement-2 is false as area of triangle formed by cube roots
of unity is 3 3 4 square units.
Example 98: Statement-1: Coefficient of xn in
n
 (-1) j (2x + 3)n – j (5 – 2x)j
j=0
is (n + 1)2n
LEVEL 2
Straight Objective Type Questions
Example 99: The number of rational terms in the ex-
(
3
6
)
pansion of 1 + 2 + 5 is
(a) 7
(b) 11
(c) 12
(d) 12
Ans. (a)
3
Solution: A term in the expansion of 1 + 2 + 5
of the form
s
r
6!
(1)6 -r - s 2 3 5
r ! s! (6 - r - s)!
(
( )( )
6
)
is
=
6!
2 r / 2 ( 5s / 3 )
r ! s! (6 - r - s)!
(
)
This term will be rational if 2r/2 and 5s/3 are both rational
numbers. This is possible if and only if r is a multiple of 2
and s is a multiple of 3. Possible values of r are 0, 2, 4 and
6 whereas the possible values of s are 0, 3 and 6. Also note
that 0 £ r + s £ 6.
For r = 0, s can take value 0, 3 or 6.
For r = 2, s can take value 0 or 3.
Mathematical Induction and Binomial Theorem
For r = 4, s can take value 0.
For r = 6, s can take value 0.
Thus, the number of rational terms is 3 + 2 + 1 + 1 = 7
(a) 3, 4
(c) 2, 7
Ans. (b)
Example 100: The sum of (n + 1) terms of the series
C0 C1 C2 C3
+
+ … is
2
3
4
5
1
1
(b)
(a)
n +1
n+2
1
1
(c)
(d)
(n + 1)(n + 2)
n(n + 1)
Ans. (d)
Solution: We have
(1 – x)n = C0 – C1 x + C2 x2 – C3 x3 + … + (–1)n Cn xn
fi
x (1 – x)n = C0 x – C1 x2 + C2 x3 – C3 x4 + … +
(–1)n Cn xn +1
fi
1
1
n
Ú x(1 – x) dx =
Ú [C0 x – C1 x
0
0
2
+ C2 x3 – C3 x4 + …
+ (– 1)n Cn xn +1]dx
(1)
L.H.S. of (1)
1
=
Ú
1
x(1 - x )n dx = Ú (1 - x ) (1 - (1 - x ))n dx
0
È
Íusing
ÍÎ
0
a
a
˘
Ú f ( x)dx = Ú f (a - x)dx ˙˙
0
˚
0
R.H.S. of (1)
(-1)n Cn n +1 ˆ ˘
x ˜˙
n +1
¯ ˚0
| x - 2|
- 9]
)
1
log7 [(4)3|x– 2| – 9] = log7 [(4) 3|x–2| – 9]1/5
5
fi
z = [(4)3|x – 2| – 9]1/5
Now, E = (y + z)7 and the 6th term in its expansion is
given by
2
T6 = 7C5 y7–5 z5 = 21 3| x - 2| (4) 3| x - 2| - 9
fi
log7 z =
(
2|x – 2|
){
}
|x–2|
fi
567 = 21{3
} {(4) 3
– 9}
3|x–2|
2|x – 2|
– (9) (3
)}
fi
27 = {(4) 3
fi
27 = (4) 33|x–2| – (9) 32|x – 2|
fi 4u3 – 9u2 – 27 = 0 where u = 3|x – 2|
Note that u = 3 satisfies this equation*.
fi
|x – 2| = 1 fi x – 2 = + 1
\
3|x – 2| = 3
x = 2 + 1 = 3 or 1.
Example 102: Let an denote the term independent of
3n
( a ) n!
sin (1 / n ) ˘
x in the expansion of È x +
, then lim 3nn
ÍÎ
x Æ•
Pn
x 2 ˙˚
equals
(b) 1
(d) e / 3
3n
fi
r=n
Ê 1ˆ
an = 3n Cn sin n Á ˜
Ë n¯
\
Ê 1ˆ
= sin n Á ˜ Æ 0 as n Æ •
3n
Ë n¯
Pn
n ! an
Example 103: The largest term in the expansion of
1
(3 + 2x)50, where x = , is
5
Example 101: Two values of x for which the sixth term
*
(
Now,
C0 C1 C2
C
+
- … + (-1)n n
n +1
2
3
4
1
=
(n + 1)(n + 2)
9 x-2
9| x - 2| = 3| x - 2|
Next, put z = 7(1 / 5) log7 [( 4 ).3
3n – r – 2r = 0
C
(-1)n Cn
C C
= 0 - 1 + 2 -…+
n +1
2
3
4
(
log3 y = log3 9| x- 2|
fi
r
È Ê 1ˆ ˘
Cr x 3n – r Ísin Á ˜ ˙ x – 2r
Î Ë n¯ ˚
For this term to be independent of x, set
2
3
4
Ê
= C0 x - C1 x + C2 x ÁË
2
3
4
of the expansion of E = 3log3
567, is
y=
Solution: Tr + 1 =
1
Thus,
fi
(c) e
Ans. (a)
1
(n + 1)(n + 2)
…+
9| x - 2 |
(a) 0
Ê x n +1 x n + 2 ˆ ˘
(
1
x
)
x
dx
=
= Ú
ÁË n + 1 - n + 2 ˜¯ ˙
˙˚0
0
n
=
(b) 3, 1
(d) none of these
log
Solution: Put y = 3 3
1
1
7.23
| x - 2|
+ 7(1 / 5) log7 [( 4 ).3
- 9]
)
7
is
(a)
50
(c)
50
2 6
C6 344 Ê ˆ
Ë 5¯
2
C43 37 Ê ˆ
Ë 5¯
(b)
50
(d)
50
43
2 7
C7 343 Ê ˆ
Ë 5¯
2
C44 36 Ê ˆ
Ë 5¯
44
Ans. (a)
From available answer, possible values of u are 3|3–2| = 3 for the choices (a) and (b) and 3|2–2| = 30 = 1 for the choice (c).
7.24
Complete Mathematics—JEE Main
Solution: Let k =
(50 + 1) (2 x )
=
(51)(2 / 5)
3 + 2x
17 / 5
As k is an integer, there are two greatest term viz. 6th and
7th terms
We have
2 6
T7 = 50 C6 344 Ê ˆ .
Ë 5¯
Remark
In fact
T 6 = T 7.
Example 104: If n is odd, then sum of the series C02 –
C1 + C22 – C23 + º + (– 1)n Cn2 is
(a) 0
(b) 1
(d) – ( 2nC n)
(c) 2n Cn
Ans. (a)
Solution: Let n = 2m – 1. Using Cr = Cn – r, we get sum
of the series in 0.
2
Example 105: If n is even, then sum of the series
C0 – C12 + C22 – C32 + º + (– 1)n Cn2 is
2
(a) n C n ( –1)n / 2
(b)
2
n
Cn
2
n
(d) ( –1)n / 2 nC2
(c) Cn – 1
Ans. (a)
2m
= Coefficient of the constant term in
r=0
(1 + x)2m Ê 1 –
Ë
Example 107: If Cr stands for nCr, the sum of the series
( )( )
n n
! !
2 2 [C 2 – 2C 2 + 3C 2 – º + (– 1)n (n + 1)C 2]
0
1
2
n
n!
where n is an even positive integer, is equal to
(a) (– 1)n/2 (n + 2)
(b) (– 1)n (n + 1)
n/2
(d) none of these
(c) (– 1) (n + 1)
Ans. (a)
2
Solution: Let n = 2m, and
S = C02 – 2C12 + 3C22 – º +
(1)
(– 1)2m (2m + 1)C22 m
Using Cr = Cn – r , we rewrite (1) as
S = (2m + 1)C02 – (2m) C12 + (2m – 1)C22 –
(2)
º + C22 m
we get
2 S = (2m + 2) [C02 – C12 + C22 – º + C 22 m]
= (2m + 2) (2m Cm) (– 1)m
fi
S = (m + 1) (2mCm) (– 1) m
n!
n
= ( –1)n / 2 Ê + 1ˆ
Ë2 ¯ n n
!
!
2 2
n n
2 ! !
2
2 S = (– 1) n/2 (n + 2)
fi
n!
( )( )
Example 108: Sum of the coefficients of integral pow-
1 ˆ 2m
x¯
50
ers of x in the expansion of (1 – 2 x )
= Coefficient of x 2m in (1 + x) 2m (x – 1) 2m
= Coefficient of x 2m in (1 – x2)2m
= 2m Cm(– 1)m = n C n ( –1)n / 2
2
n
Example 106: If (1 + x) = C0 + C1 x + C2 x2 +
º + Cn xn, then value of C02 + 2C12 + 3C22 + º +
(n + 1) C2n is
(b) (2n – 1) (2nCn)
(a) (2n + 1) (2nCn)
n
(c) Ê + 1ˆ ( 2 n Cn )
Ë2 ¯
Ans. (c)
fi
( )( )
Solution: Let n = 2m. We have
 (–1)r Cr2
2S = (n + 2) [C02 + C12 + º + Cn2] = (n + 2) (2nCn)
n
S = Ê + 1ˆ ( 2 n Cn )
Ë2 ¯
=6.
n
(d) Ê1 + ˆ ( 2 n – 1 Cn )
Ë 2¯
Solution: Let
S = C02 + 2C12 + 3C22 + º + (n + 1)Cn2
Using Cr = Cn – r , we rewrite (1) as
S = (n + 1)C 02 + nC12 + º + Cn2
Adding (1) and (2), we get
(1)
(2)
1 50
(3 + 1)
2
1
(c) (250 – 2 )
2
Ans. (a)
(a)
is
1 50
(3 )
2
1
(d) (250 + 2 )
2
(b)
Solution: The coefficients of integral powers of x are
50
C0, 50C2 22, 50C4 24, º, 50C50 250
Let
S = 50C0 + 50C2 22 + 50C4 24 + º + 50C50 250
1
1
= ÈÎ(1 + 2 )50 + (1 – 2 )50 ˘˚ = [350 + 1]
2
2
Example 109: Given that the 4th term in the expansion
3 x ˆ 10
Ê
of Á 2 + ˜ has the maximum numerical value, then x
Ë
8¯
lies in the interval
Ê 64
ˆ Ê 64 ˆ
(a) Á - , - 2˜ » Á 2, ˜
Ë 21
¯ Ë 21 ¯
Mathematical Induction and Binomial Theorem
Example 111: If (1 + x + x2)8 = a0 + a1x + a2 x2 +
a16 x16, then a5 equals
(a) 504
(b) 506
(c) 508
(d) 502
Ê 60
ˆ Ê 64 ˆ
(b) Á - , -2˜ » Á 2, ˜
Ë 23
¯ Ë 21 ¯
Ê 64
ˆ
(c) Á - , - 2˜
Ë 21
¯
(d) none of these
Ans. (a)
(1 + x + x2)8 =
fi
|T3| < |T4| and |T5| < |T4|
T3
T
< 1 and 5 < 1.
T4
T4
But
T3
=
T4
2
C 2 ( 28 ) (3 x 8 )
3
7
=
C3 2 (3 x 8)
and
T5
=
T4
Now,
T3
<1
T4
fi
2
<1
x
2
x
Next,
T5
<1
T4
fi
21x
<1
64
fi
x < – 2 or x > 2.
fi -
64
64
<x<
21
21
Ê 64 64 ˆ
Thus x Œ (-•, -2) » (2, •) and x ŒÁ - , ˜
Ë 21 21 ¯
Ê 64
ˆ Ê 64 ˆ
Hence, x ŒÁ - , -2˜ » Á 2, ˜ .
Ë 21
¯ Ë 21 ¯
Example 110: Assuming x to be so small that x3 and
higher powers of x can be neglected, then value of
3 ˆ5
Ê
6
E = Á 1 - x˜ (2 + 3 x ) , is
Ë 2 ¯
(a) 64 + 96x + 720 x2
(c) 64 – 96x + 720x2
Ans. (d)
(b) 65 – 97x + 721x2
(d) 64 + 96x – 720x2
Solution: We have
3 ˆ5
3 ˆ5 Ê
3 ˆ6
Ê
6Ê
6
1
x
(
2
+
3
x
)
=
2
1
x
1
+
x˜
ÁË
˜
ÁË
˜ Á
2 ¯
2 ¯ Ë 2 ¯
3 ˆ ÈÊ
3 ˆÊ
3 ˆ˘
Ê
= 26 Á 1 + x˜ ÍÁ 1 - x˜ Á 1 + x˜ ˙
Ë 2 ¯ ÎË 2 ¯ Ë 2 ¯ ˚
3 ˆ
Ê
= 2 Á 1 + x˜
Ë 2 ¯
9 2ˆ5
Ê
1
x ˜
ÁË
4 ¯
3 ˆ
Ê
= 26 Á 1 + x ˜
Ë 2 ¯
45 2 ˆ
Ê
x ˜
ÁË 1 4 ¯
6
Â
p, q ≥ 0
p+q£8
8!
x p (x2)q
p! q! (8 - p - q )!
For coefficient of x5, we set p + 2q = 5
This is possible if p = 5, q = 0, p = 3, q = 1, p = 1, q = 2
Thus, coefficient of x5 is
8!
8!
8!
+
= 504
+
5! 3!
3! 4! 2! 5!
( )
4
10
C 4 ( 2 6 ) (3 x 8 )
21x
=
3
10
7
64
C3 (2 ) (3 x 8)
10
+
Solution: We have
Solution: Since T4 is numerically the greatest term,
10
7.25
Example 112: Coefficient of x11 in the expansion of
(1 + x2)4 (1 + x3)7 (1 + x4)12 is
(a) 1051
(b) 1106
(c) 1113
(d) 1120
Ans. (c)
Solution: We have
(1 + x2)4 (1 + x3)7 (1 + x4)12
= (1 + x2)4 (1 + 7x3 + 31x6
+ 35x9 + …) (1 + 12x4 + 66x8 + …)
2 4
As expansion of (1 + x ) contains only even power, while
multiplying last two expansions, we shall only consider odd
powers.
\ (1 + x2)4 (1 + x3)7 (1 + x4)12
= (1 + 4x2 + 6x4 + 4x6 + x8) (7x3 + 84x7
+ 35x9 + 462x11 + …)
11
Thus, coefficient of x is
(1) (462) + 4(35) + (6) (84) + (1) (7) = 1113
Example 113:
terms of (1 + x)n + 5 are in the ratio 5:10:14, then n is equal to
(a) 5
(b) 6
(c) 7
(d) 8
Solution: Let three consecutive coefficients of (1 + x)n + 5
be n + 5Cr, n + 5Cr + 1, n + 5Cr + 2
We are given
n +5
5
3
45 2 ˆ
Ê
x ˜ = 64 + 96 x - 720 x 2
= 26 Á 1 + x Ë 2
4 ¯
n +5
fi
fi
\
Cr
Cr +1
=
5
and
10
n +5
n +5
Cr +1
Cr + 2
=
10
14
1
r +1
r+2
5
= and
=
n+5-r 2
n + 5 - r -1 7
3r = n + 3 and 12r = 5n + 6
4(n + 3) = 5n + 6 fi n = 6
Example 114: Sum of the series
n
 n Ck ( -1)k
k =0
1
ak
Complete Mathematics—JEE Main
7.26
where
ak =
m
k
= Â ( -1) k ( m Ck )
 kCi bi
k =0
i =0
i
Ê -2 ˆ
bi = Â iC j Á ˜
Ë 3¯
j =0
where
j
Thus,
(c)
1
2
(b)
n
1
1
3n
Ê 3ˆ
(d) Á ˜
Ë 4¯
4n
S–T=0
Example 116: Coefficient of xm in the expansion of
S = (1 + x)2m + x(1 + x)2m – 1 + x2(1 + x)2m – 2
+ …. + x2m
is
(a)
is
n
(a) 2mCm + 2m – 1Cm – 1
(b)
2m + 1
Cm
2
(c) 0
Ans. (c)
(d) 2m + 2m + 1
Solution: Note S is a G.P. with first term (1 + x)2m and
Solution:
j
i
i
Ê -2 ˆ
Ê 2ˆ Ê 1ˆ
bi = Â i C j Á ˜ = Á1 - ˜ = Á ˜
Ë
¯
Ë 3 ¯ Ë 3¯
3
j =0
i
k
k
1
1
4
ak =  k Ci ÊÁ ˆ˜ = ÊÁ1 + ˆ˜ = ÊÁ ˆ˜
Ë 3¯ Ë 3¯
Ë 3¯
i =0
\
1
=T
n + k +1
n
k
i
common ratio
Thus,
È Ê x ˆ 2 m +1 ˘
(1 + x)2 m Í1 - Á
˙
˜
˙˚
ÍÎ Ë 1 + x ¯
S=
1 - x / (1 + x)
k
n
1
Ê 3ˆ
Ê 3ˆ
 Ck ( -1)k ÁË ˜¯ = ÁË1 - ˜¯ = n
4
4
4
k =0
Thus,
n
= (1 + x)2m + 1 – x2m + 1
xm in S is 2m + 1Cm.
Example 115: Suppose m and n
and let
n
S = Â ( -1)
k
k =0
m
1
( n Ck )
k + m +1
1
and
T = Â ( -1)
( m Ck ) ,
k
n
+
+
1
k =0
then S – T is equal to
(a) 0
(b) nm – mn
k
(c) (n + 1)m – (m + 1)n
Ans. (a)
(d) (1 – n)m – (1 – m)n
Solution:
n
Ê 1 k +m ˆ
k n
S = Â ( -1) ( Ck ) Á Ú0 x dx˜
Ë
¯
k =0
n
ˆ
1Ê
= Ú0 Á Â ( -1) k ( n Ck )x k + m ˜ dx
Ë k =0
¯
1
= Ú0 (1 - x)n x m dx
1
= Ú0 (1 - (1 - x)) n (1 - x)m dx
x
and number of terms 2m + 1.
1+ x
Example 117: Sum of the first 20 terms of the series
(1)(3)
(1)(3)(5)
1
+
+
+…
(2)(4) (2)(4)(6) (2)(4)(6)(8)
is
(a)
1
1
- 40 ( 40 C20 )
2 2
(b)
1 1 42
( C21 )
2 241
(c)
1
1
- 42 ( 42 C21 )
2 2
(d)
1
1
- 43 ( 40 C20 )
2 2
Ans. (c)
Solution: Let ak denote the kth term of the series, then
(1)(3)(5) (2k - 1)
ak =
(2)(4)(6) (2k )(2k + 2)
(1)(3)(5) (2k - 1)(2k + 2 - 2k - 1)
=
(2)(4)(6) (2k )(2k + 2)
= bk – bk + 1
where
(1)(3)(5) (2k - 1)
bk =
(2)(4)(6) (2k )
1
= Ú0 x n (1 - x)m dx
=
Ê m
ˆ
1
= Ú x n Á Â ( -1) k ( m Ck ) x k ˜ dx
0
Ë k =0
¯
=
m
1
= Â ( -1) k ( m Ck ) Ú x n + k dx
0
k =0
(1)(2)(3)(4) (2k - 1)(2k )
[(2)(4)(6) (2k )]2
1
22 k
( 2 k Ck )
Thus,
20
20
 ak =  (bk - bk +1 ) = b1 – b21
k =1
k =1
Mathematical Induction and Binomial Theorem
=
7.27
n
1
1
- 42 (42C21)
2 2
= Â ak [1 – (1 – x)]k
k =0
n
Ê k
k k
jˆ
= Â ak Á Â (-1) ( C j )(1 - x) ˜
Ë j =0
¯
k =0
Example 118: The term independent of x in the expansion of
1
1ˆ
Ê
ˆÊ
+ 3x7 ˜ Á x 2 + ˜
ÁË1 ¯
Ë
2x
3x ¯
10
n
= Â bk (1 - x) k
k =0
is
n
(a) 10/38
(c) 10/39
Ans. (a)
is
j =k
1ˆ
Ê
Solution: (r + 1)th term in the expansion of Á x 2 + ˜
Ë
3x ¯
Ê 1ˆ
tr + 1 = 10Cr (x2)10 – r Á ˜
Ë 3x ¯
k j
bk = Â ( -1) ( Ck )a j
where
(b) – 2/3
(d) 11/6
10
Ê n
ˆ
1
1
Now Ú0 ( P( x)) 2 dx = Ú0 P( x) Á Â bk (1 - x) k ˜ dx
Ë k =0
¯
n
r
k =0
As P(x) is a polynomial,
P(x) ∫ 0 " x Œ [0, 1]
fi
P(x) ∫ 0
r
Ê 1ˆ
= 10Cr Á ˜ x20 – 3r
Ë 3¯
We require terms containing x0, x and x–7 in the expansion
Example 120: Suppose [x] denote the greatest integer
£ x, and n Œ N, then
10
1ˆ
Ê
of Á x 2 + ˜ .
Ë
3x ¯
lim
[n Co x 2 ] + [n C1 x 2 ] + + [ n Cn x 2 ]
2n - 2
nƕ
0
For x , set 20 – 3r
For x1, set 20 – 3r
For x–7, set 20 – 3r = –7
r.
r.
is equal to
1 2
x
(a)
2
r = 9.
n
Ú
P(1 – x)dx = 0 " m
n
fi
n
(a) P(x) = x (1 – x) for some n N
(b) P(x) = (1 – x)2n for some n N
(c) P(x) = 1 – xm (1 – x)n for some m, n
(d) P(x) = 0
Ans. (d)
Solution: 0 =
{0}, then
N
1 m
Ú0 x
1
k
P(x) = Â ak x where ak
k =0
fi
n
n
k =0
k =0
k =0
x 2 (2n ) - (n + 1)
2n - 2
<
£
m
Taking limit as n
0
lim
R
n
nƕ
R, therefore
 (n Ck x 2 - 1) <  [n Ck x 2 ] £  n Ck x 2
N
P (1 - x ) dx = Ú (1 - x ) P ( x )
n
(d) 4x2
Solution: We know x – 1 < [x] £ x " x
Ck x2 – 1 < [nCk x2] £ nCk x2
Example 119: Let P(x) be a polynomial with real coef1 m
x
0
(b) x2
(c) 2x2
Ans. (b)
9
È
Ê 1 ˆ ˘ 10
3 Í 10 C9 Á ˜ ˙ = 8
Ë 3¯ ˙ 3
ÍÎ
˚
Let
1
= Â bk Ú0 (1 - x)k P( x)dx = 0
1
n
1
2n - 2 k = 0
x 2 (2n )
2n - 2
•, and using sandwich theorem, we get
 ÈÎ n Ck x 2 ˘˚ = x 2
2n - 2 k = 0
n
 [ n Ck x 2 ]
7.28
Complete Mathematics—JEE Main
EXERCISE
Concept-based
Straight Objective Type Questions
1. Suppose k Œ R. If the coefficient of x in the expankˆ5
Ê
sion of Ë x 2 + ¯ is 270, then k is equal to:
x
(a) 3
(b) – 3
(c) 9
(d) – 9
2. Suppose p ΠR. If the fourth term in the expansion
1ˆ n
5
Ê
of Ë px + ¯ is , then (n, p) is equal to:
x
2
(a) (5, 1/2)
(b) (6, 1/2)
(c) (8, 1/2)
(d) (10, 1/2)
3. If the coefficient of (3r
term in the expansion of
equal to:
(a) 4
(c) 7
+ 4)th term and (2r – 2)th
(1 + x)20 are equal then r is
(b) 5
(d) 8
4. If sum of the coefficient in the binomial expansion of
(1 + 2x)n is 6561, then coefficient of the fourth term
is:
(a) 442
(b) 446
(c) 448
(d) 472
5. Coefficient of x10 in the expansion of
2 10
(1 + x )
(a) 10C5
(c) 0
1ˆ
Ê 2
ÁË x + 2 + 2 ˜¯
x
6. If (1 + ax)n = 1 + 10x + 40x2 + . . . , then value of
a+n
is
a-n
(a) 5/7
(b) 7/3
(c) –5/7
(d) – 7/3
7. The ratio of the coefficient of x10 in (1 + x2)10 and
2 ˆ 10
Ê
the term independent of x in Ë x + ¯ is
x
(a) 32 : 1
(b) 1 : 32
(c) 1 : 64
(d) 64 : 1
8. If coefficient of 9th, 10th and 11th terms of (1 + x)n
are in A.P., then value of n can be
(a) 15
(b) 32
(c) 23
(d) 17
9. (p + 2)th term from the end in the binomial expansion
2 ˆ 2n +1
Ê
is
of Á x 2 - 2 ˜
Ë
x ¯
(a) 2n + 1C2n – p (– 2)2n – p x2p + 1 – 2n
(b) 2n + 1C2n – p (– 2)2n – p x2n – 2p
(c) 2n + 1C2n – p (– 2)2n – p x2n – 2p + 1
(d) 2n + 1C2n – p (– 2)2n – p x2p – 1 + 2n
10. Let Cr =
5
15
is
S=
(b) 15C5
(d) 2
15
Cr, 0 £ r £ 15. Sum of the series
Ê C ˆ
˜ is
r -1 ¯
 r ÁË C r
r =1
(a) 40
(c) 100
(b) 60
(d) 120
LEVEL 1
Straight Objective Type Questions
1ˆ n
Ê
11. If the middle term of Ë x 2 + ¯ is 924 x6, then value
x
of n is
(a) 8
(b) 10
(c) 12
(d) 20
1 ˆ 20
Ê
12. The greatest term in the expansion of 5 Á 1 +
˜
Ë
5¯
is:
Ê 1ˆ
Ê 1ˆ
(b) 20C5 Ë ¯
(a) 20C5 Ë ¯
25
5
20
20
(c) C5
(d) C5 (5)
13. Let S(q ) denote the sum of coefficients in the expansion of ( 2 – x sin q + x2 cos q )2n. Maximum value
of S(q ) is
(a) 4n
(b) 8n
n
(c) 2
(d) 1
14. Suppose
(x + a)n = T0 + T1 + T2 + . . . + Tn
Then (T0 – T2 + T4 – . . .)2 + (T1 – T3 + T5 – . . .)2
is equal to
(b) (x2 – a2)n
(a) (x2 + a2)n
2n
(c) (x + a)
(d) (x – a)2n
Mathematical Induction and Binomial Theorem
15. For 0 £ r < 2n, (2n + rCn) (2n – rCn) cannot exceed
(b) 4nC2n
(a) 4nCn
6n
(c) C3n
(d) none of these.
16. If sum of the coefficients in the expansion (2x2 – 3cx
+ c2)17 is zero, then c is equal to
(a) 2, 3
(b) 1, 2
(c) 1, 3
(d) 2, – 1
17. Coefficient of constant term in the expansion of
3
(x3 + 5(3) – log 3 x )4 is
(a) 125
(b) log
(c) log3 5
(d) 150
3
5
18. If the number of terms in the expansion of
(1 + 5x + 10x2 + 10x3 + 5x4 + x5)20 is m, then unit’s
place of 2 m is
(a) 2
(b) 8
(c) 6
(d) 4
19. Let Pn denote the product of all the coefficients in the
expansion of (1 + x)n. If (20)! Pn + 1 = 2120 Pn, then
n is equal to
(a) 21
(b) 20
(c) 19
(d) 18
20. Coefficient of the term independent of x in the expan-2 n
Ê x ˆ
is:
sion of (1 + x)2n Á
˜
Ë1 - x¯
(b) (–1)n 2nCn
(a) 2nCn
n2 – n
(c)
Cn
(d) 0
10
Êx 3ˆ
4
21. Coefficient of x in the expansion of Á - 2 ˜ is
Ë2 x ¯
504
(a)
256
(c)
459
512
405
(b)
256
(d) -
135
256
22. If a and b are the coefficients of xr and xn– r respectively in the expansion of (1 + x)n, then
(a) a = b
(b) a + b = n
(c) a = kb for some constant k
(d) a + b = 2n
23. If the coefficients of (2r + 4)th and (r – 2)th terms in
the expansion of (1 + x)18 are equal, then value of r is
(a) 5
(b) 6
(c) 7
(d) 9
24. The total number of terms in the expansion of
(x + a)200 + (x – a)200 after simplification is
(a) 101
(b) 102
(c) 201
(d) 202
7.29
25. The coefficient of x5 in the expansion of
(1 + x2)5 (1 + x)4 is
(a) 30
(b) 40
(c) 50
(d) 60
26. If n > 2, r > 1 and the coefficients of (3r)th and
(r + 2)th terms in the expansion of (1 + x)2n are
equal, then
(a) n = 2r
(b) n = 3r
(c) n = 2r + 1
(d) n = 2r + 2
27. If (1 + px)n = 1 + 24x + 2642x2 +… then
(a) p = 2, n = 6
(b) p = 2, n = 12
(c) p = 3, n = 6
(d) none of these
28. Sum of the series
(a)
(c)
10
È1
r =0
Î
3r ˘
 (-1)r ( 10 Cr ) Í 2r + 22r ˙
210 + 1
210 - 1
(b)
220
is
˚
220
220 - 1
(d) none of these
220
29. If m and n are positive integers, then value of
m
C0 nCk + mC1 nCk –1 + …+ mCk nC0
(a) m +nCk
(b) m+nCm+k
m+n
(c)
Cn+k
(d) none of these
30. The expression
(x +
polynomial of degree
(a) 6
(c) 9
x3 - 1
6
) + (x -
x3 - 1
6
)
is a
(b) 7
(d) 10
31. The number of distinct terms in the expansion of (1
+ 3x + 3x2 + x3)7 is
(a) 18
(b) 19
(c) 28
(d) 22
32. If x > 0, then the number of positive terms in the
expansion of (1 + ix)4n is
(a) n + 1
(b) n – 1
(c) n
(d) 2n
33. The number of terms whose values depend on x in
1 ˆn
Ê
the expansion of Á x 2 - 2 + 2 ˜ is
Ë
x ¯
(a) 2n
(c) 2n – 1
(b) 2n + 1
(d) 0
5
Ê1
ˆ
34. If the third term in the expansion of Á + x log10 x ˜ ,
Ëx
¯
x > 1, is 1000, then value of x is
(a) 10
(c) 100
(b) 10
(d) 10 10
7.30
Complete Mathematics—JEE Main
35. Let R =
(
2 n +1
)
2 +1
, n ΠN, and f = R Р[R], where [ ]
denote the greatest integer function, Rf is equal to
(a) 1
(b) 22n +1
2n
(c) 2 – 1
(d) none of these
36. The coefficient of x 60 in (1 + x)51 (1 – x + x 2) 50 is
(b) – (50C20)
(a) 50C20
51
(d) – ( 51C20)
(c) C20
37. If a = 9950 + 10050 and b = 10150, then
(a) a < b
(b) a = b
(c) a > b
(d) a – b = 10049
38. If (1 + x – 2x ) =
 Ar x ,
r
then value of A2 + A4
(b) 31
(d) none of these
(
6
)
3 +1
is
41. The number of terms in the expansion of (x2 + 6x +
9)30 is
(a) 31
(b) 61
(c) 91
(d) none of these
42. The expansion of (x + y + z)n is given by
n!
(a) Â
x p y q z r where p, q, r ≥ 0, p + q +r = n
p! q ! r !
(c)
n!
(a) 1
(c) nx
1ˆ n
is
x¯
(a) 51
(b) 48
(c) 35
(d) 28
48. If, in the binomial expansion of (a – b)n, n ≥ 5 the sum
a
of 5th and 6th term is zero, then value of
is
b
n-5
6
(a)
(b)
6
n-4
n-4
5
(c)
(d)
5
n-4
49. The value of 2C0 +
(d) none of these
43. The number of terms in the expansion of the multinomial (x + y + z)n is
1
1
(a)
(n + 1) (n + 2)
(b)
n(n + 1)
2
2
1
(n + 1) (n + 3)
(d) none of these
2
44. The number of terms in the expansion of (x1 + x2
+…+ xr)n is
(c)
(c) n +r +1Cr +1
(d)
n+r– 1
n + r –1
1
(211 – 1)
11
(b)
1
(311 – 1)
11
(c)
1
(113 – 1)
11
(d)
1
(112 – 1)
11
50. If Cr = nCr, then value of
C3
Cn ˆ
C2
Ê C1
2 ÁË C + 2 C + 3 C + n C ˜¯ is
0
1
n -1
2
(a) n(n –1)
(b) n(n + 1)
2
(d) n2 + 1
(c) n – 1
n
51. If Cr = Cr , then the value of
Cn ˆ
C1 ˆ Ê
C2 ˆ Ê
Ê
ÁË 1 + C ˜¯ ÁË 1 + C ˜¯ ÁË 1 + C
˜
n –1¯
0
1
(a)
(n + 1)!
2n
Cr – 1
Cr
22
23
211
C1 + C2 + … +
C10 , is
2
3
11
(a)
(n + 1)!
 p! q!r ! x p yq zr , where p + q + r = n
(b)
(b) 0
(d) ny
Ê
Ëx +
where p + q + r = n
(a) n +1Cr
is
1
47. If sum of the coefficients in the expansion of Ê x + ˆ
Ë
x¯
is 128, then coefficient of x in the expansion of
(b) 416
(d) 207
 p! q ! r ! x p y q z r ,
 (r ) (n Cr ) x n-r yr
n
40. The greatest integer contained in
(b)
(b) ab = 11
(d) ab = 6
r =0
39. The sum of the coefficients in the expansion of (1 +
x – 3x2)4321 is
(a) 0
(b) 1
(c) –1
(d) 24320
(a) 208
(c) 415
(a) ab = 1
(c) ab = 5
n
r =0
+ A6 +…+ A12, is
(a) 30
(c) 32
1 ˆ 11
Ê
ax
˜ are equal, then
ÁË
bx 2 ¯
46. If x + y = 1, then value of
12
2 6
11
1ˆ
Ê
45. If the coefficient of x7 in the expansion of Á ax 2 + ˜
Ë
bx ¯
– 7
is the expansion of
and the coefficient of x
(c)
2
(b)
n –1
n!
(n + 1)n
n!
n
(d)
2 –1
n!
Mathematical Induction and Binomial Theorem
1
1
52. If Cr = nCr , then C0 – C1 + C2 upto (n + 1)
3
5
terms equal
(a)
1
Ú0 x (1 – x )
n
dx
(b)
1
Ú0 (1 – x 2 )
n
dx
ˆ
Ê 2
sion of Á x – b ˜
Ë a x¯
(a) ab = 1
(c) ab = –1
1
(d) 1 + Ú x (1 + x )n dx
(c) 0
0
60. The value of
n+4
53. Let (1 + x2)2 (1 + x)n =
 ar xr . If a1, a2, a3 are in
r =1
A.P., then n is equal to
(a) 6
(c) 7
(b) 5
(d) 2, 3, 4
n
54. The value of
 ( 4 n+1Ck + 4 n+1 C2 n- k )
is
k =0
(a) 24n + 4n +1C2n
(c) 24n
(b) 24n +1
(d) 24n+1 + 4n +1Cn
55. Coeffiecient of t 24 in the expansion of
(1 + t 2) 12 (1 + t 12) (1 + t24) is
(b) 12C6 + 1
(a) 12C6 + 2
12
(d) 12C6 + 3
(c) C6
56. If the value of the third term in the expansion of
5
( x + x log x ) is 106, then x may have value(s)
10
(b) 10, 10–5/2
(d) 102, 10–3/2
(a) 1, 10
(c) 10, 10–3/2
57. The greatest term in the expansion of (1 + x)10 when
x = 2/3 is
(a) 210(3/2)6
(b) 210(2/3)4
4
(c) 210(3/2)
(d) none of these
58. The greatest term in the expansion of (3 + 5x)15, when
x = 1/5, is
(b) 15C4 (312)
(a) 15C3 (313)
15
10
(d) none of these
(c) C4 (3 )
59. If sum of the coefficients of x7 and x4 in the expan-
7.31
11
is zero, then
(b) a = b
(d) a + b = 0
193 + 63 + (3)(19)(6)(25)
36 + 6(243)(2) + (15)(81)(4) + (20)(27)
(8) + (15)(9)(16) + (6)(3)(32) + 26
(a) 19
(c) 25
is
(b) 6
(d) 1
61. In the expansion of (x + a)n, if the sum of odd terms
be A and the sum of even terms be B, then value of
(x2 – a2)n is
(a) AB
(b) 2AB
(c) 4AB
(d) A2 – B2
62. Coefficient of the term independent of x in the expan1ˆ 4 Ê
1 ˆ 12
Ê
sion of Á x + ˜ Á x - ˜ is
Ë
x¯ Ë
x¯
(a) 192
(c) 196
(b) 194
(d) 198
63. For all n ΠN, (3) (52n +1) + 23n +1 is divisible by
(a) 19
(b) 17
(c) 23
(d) 25
64. 1.2.3 + 2.3.4 + 3.4.5 + … upto n terms is equal to
1
(n + 1) (n + 2) (n + 3)
(a)
4
1
(n + 2) (n – 2) (n – 3) (n + 3)
(b)
4
1
(c)
n(n + 1) (n + 2) (n + 3)
4
(d) none of these
65. If a, b,c, ΠN, an + bn is divisible by c when n is
odd but not when n is even, then value of c is
(a) a + b
(b) a – b
(d) a3 + b3
(c) a3 + b3
Assertion-Reason Type Questions
66. Let (1 + x)n = C0 + C1x + C2 x 2 + º + Cnxn
n
Statement-1:
= 2n sin (nx)
Â
Cr sin (rx) cos (n – r)x
r=0
n
Statement-2:
Â
r=0
Cr = 2 n
67. Statement-1: The coefficient of the term of inden
9
Ê
ˆ
pendent of x in the expansion of Á x + + 6˜ is
Ë
¯
x
3n (2n )!
.
n! n!
7.32
Complete Mathematics—JEE Main
Statement-2: The coefficient of x r in the expansion
Ê nˆ
of (1 + x) n is Á ˜ .
Ë r¯
68. Statement-1: For any positive integers m, n (with
n ≥ m),
Ê n ˆ Ê n – 1ˆ Ê n – 2ˆ
Ê mˆ Ê n + 1ˆ
(1)
ÁË m˜¯ + ÁË m ˜¯ + ÁË m ˜¯ + + ÁË m˜¯ = ÁË m ˜¯
Statement-2: If p is a prime and 1 £ r £ p – 1,
Ê pˆ
then Á ˜ is divisible by p.
Ë r¯
70. Statement-1: Greatest term in the expansion of
(
(
integer contained in 2 + 5
by 20 n.
)
–2
is
È ( n + 1) x ˘
integer and r = Í
˙ , where [y] denotes the
Î x +1 ˚
greatest integer £ y.
69. Statement-1: If n is an odd prime, then greatest
n +1
50
)
Ê 50ˆ 14 11
ÁË 22˜¯ 3 2 .
Statement-2: Greatest term in the expansion of
(n + 1) x
(1 + x) n, x > 0 is the rth term if
is not an
x +1
Statement-2: Coefficient of x r in the expansion of
Ê nˆ
(1 + x)n is Á ˜ .
Ë r¯
n
3+ 2
is divisible
LEVEL 2
Straight Objective Type Questions
71. Let S denote the set of real numbers x such that
(x +
x
2
3
- 1) + ( x -
Then S contains
(a) one element
(c) infinte elements
x
2
- 1)
3
= 2cos (3cos
(b) two elements
(d) no element
72. Value of
(40C0) (40C15) – (40C1)(40C16) + (40C2)(40C12)
– (40C25) (40C40)
equals
(a) 0
(b) 40C25
40
(c) C20
(d) –1
73. The hundred’s digit of 3100 is
(a) 0
(b) 1
(c) 2
(d) 7
1
p(p – 1)
2
(d)
1
n
1
(c)
n
( 2 n Cn )
(b)
(a) nn + 2nCn
(c) 2nCn
77. Sum of the series
ÊC ˆ
S = n2 Á 0 ˜ + (n – 1)2
Ë C1 ¯
equals
1
(a)
n(n+1) (2n + 1)
6
1
n (n + 1) (n + 2)
(c)
6
1
n+k
Cn
Ê Cn - 1 ˆ
+12 Á
Ë Cn ˜¯
Ê C1 ˆ
ÁË C ˜¯ +
2
1
n (n + 1)(n + 2)
3
1 2
(d)
n (n + 1) (n + 2)
6
(b)
2n
r
 ar ( x - 2) =
ak =
( 2 n + 1 Cn )
 br ( x - 3)r
and
r=0
for 0 £ k £ n then bn equals
(a) n – 1
(c) 2n – 1
(b) n
(d) n + 1
79. Coefficient of x50 in the expansion of
(1 + x)41 (1 – x + x 2)40
is
(a) 0
(b) 1
(d) 40C29
(c) 40C19
80. If n ΠN, then
 ( n Ci )( n + 1C j )
equals
i< j
( 2 n Cn + 1 )
(d) 2n
equals
(b) (2nCn)2
(d) nn
r=0
1
p(p + 1)
2
1
n +1
x)
(2n )!
 (k !)2 (n - k !)2
k =0
2n
1
1
1
75. C02 + C12 + C22 + +
Cn2 equals
n +1
2
3
(a)
–1
78. If
74. If a,b, p ΠN and p is prime, then
(a + b)p – ap – bp is divisible by
(a) p
(b) p2
(c)
n
76. Sum of the series
(a) 22n
(c) 2n + 1Cn
(b) 22n – 1
(d) 2n + 1
Mathematical Induction and Binomial Theorem
7.33
Previous Years' AIEEE/JEE Main Questions*
1. If sum of the coefficients in the expansion of
(x + y) n is 4096, then the greatest coefficient in the
expansion is
(a) 792
(b) 924
(c) 1594
(d) 2990
[2002]
2. Let a r denote the coefficient of x r in the expansion
of (1 + x) p + q, then
(b) ap = – aq
(a) ap = aq
(c) ap aq = 1
(d) none of these [2002]
3. The positive integer just greater that (1 + 0.0001)10000
is
(a) 2
(b) 3
(c) 4
(d) 5
[2002]
4. If r, n ΠN r > 1, n > 2 and the coefficient of (r +
2)th term and (3r)th term in the expansion of (1 +
x)2n are equal, then n equals
(a) 3r
(b) 3 r + 1
(c) 2r
(d) 2 r + 1
[2002]
5. The number of integral terms in the expansion of (31/2
+ 51/8)256 is
(a) 33
(b) 34
(c) 35
(d) 32
[2003]
6. *If x is positive, the first negative term in the expansion of (1 + x)27/5 is
(a) 5th term
(b) 8th term
(c) 6th term
(d) 7th term
[2003]
7. Let S(K): 1 + 3 + 5 + º + (2K – 1) = 3 + K 2.
Then which of the following is true?
(a) S(K) fi
/ S(K + 1)
(b) S(K) fi S(K + 1)
(c) S(1) is correct
(d) Principle of mathematical induction can be used to
prove the formula.
[2004]
8. The coefficients of the middle terms in the binomial expansions in powers of x of (1 + a x)4 and of
(1 – ax)6 is the same if a equals
(a) – 3
10
5
(c) –
3
10
(b)
3
(d)
3
5
9. The coefficient of x n in the expansion of
(1 + x) (1 – x) n is
(b) (– 1)n (1 – n)
(a) (– 1)n – 1 (n – 1)2
(c) n – 1
(d) (– 1)n – 1 n
[2004]
10. If the coefficients of r th, (r + 1)th and (r + 2)th terms
in the binomial expansion of (1 + y) m are in A.P., then
m and r satisfy the equation
(a) m 2 – m(4r + 1) + 4r2 – 2 = 0
(b) m 2 – m(4r – 1) + 4r2 + 2 = 0
(c) m 2 – m(4r – 1) + 4r2 – 2 = 0
[2005]
(d) m2 – m(4r + 1) + 4r 2 + 2 = 0
1 11
11. If the coefficient of x7 in ÍÈax 2 + ˙˘ , equals the coÎ
bx ˚
11
1
efficients of x–7 in ÈÍax – 2 ˘˙ , then a and b satisfy
Î
bx ˚
the relation
a
=1
(b) ab = 1
(a)
b
(c) a – b = 1
(d) a + b = 1
[2005]
3
12. *If x is small that x are higher powers of x may be
neglected, then
(
1
(1 + x )3 2 – 1 + x
2
)
3
(1 – x )1 2
may be approximated as
3
(a) – x 2
8
3
(c) 1 – x 2
8
1
3
x – x2
2
8
3
(d) 3 x + x 2
8
(b)
[2005]
13. For natural numbers m and n, if (1 – y) m (1 + y)n
= 1 + a 1y + a 2 y 2 + º
and a1 = a2 = 10, then (m, n) equals
(a) (35, 45)
(b) (20, 45)
(c) (35, 20)
(d) (45, 35)
[2006]
14. The sum of the series
20
[2004]
C0 – 20C1 + 20C2 – 20C3 + º + 20C10 is
1
(a) – ( 20 C10 )
(b) ( 20 C10 )
2
(c) 0
* Questions marked with * may be skipped for the present JEE Main syllabus.
(d)
20
C10
[2007]
7.34
Complete Mathematics—JEE Main
15. In the binomial expansion of
(a – b) n, n ≥ 0, the sum of 5th and 6th terms is zero,
a
then
equals
b
5
6
(a)
(b)
n–4
n–5
(c)
n–5
6
(d)
n–4
5
[2007]
25. The ratio of the coefficient of x15 to the term inde2 ˆ 15
Ê
pendent of x in the expansion of Ë x 2 + ¯ is:
x
(a) 7 : 16
(b) 7 : 64
(c) 1 : 4
(d) 1 : 32 [2013, online]
n
16. Statement-1:
Â
(r + 1) ( nC r) = (n + 2)2n – 1
r=0
n
Statement-2:
Â
(r + 1) ( nCr) x r
r=0
n
= (1 + x) + nx (1 + x)n – 1
[2008]
17. The remainder left out when 82n –622n+1 is divided by
9 is
(a) 7
(b) 8
(c) 0
(d) 2
[2009]
10
10
j =1
j =1
18. Let S1 = Â j ( j - 1) (10 C j ), S2 = Â j (10 C j )
and
10
S3 =
 ( j 2 ) (10 C j )
j =1
Statement-1: S3 = 55 ¥ 29
Statement-2: S1 = 90 ¥ 28 and S2 = 10 ¥ 28
[2010]
19. The coefficient of x7 in the expansion of (1 – x – x2
+ x3)6 is
(a) 132
(b) 144
(c) – 132
(d) – 144
[2011]
20. Statement-1: For each natural number n,
(n + 1)7 – n7 – 1 is divisible by 7.
Statement-2: For each natural number n, n7 – n is
divisible by 7.
[2011]
(
2n
)
(
(b) e
(a) e2
1
(c) e
(d) 2e
[2013, online]
2
24. If for positive integers r > 1, n > 2, the coefficients of
the (3r)th and (r + 2)th powers of x in the expansion
of (1 + x)2n are equal, then n is equal to:
(a) 2r
(b) 2r – 1
(c) 3r
(d) r + 1
[2013, online]
2n
)
21. If n is a positive integer, then 3 + 1 - 3 - 1
is
(a) an odd positive integer
(b) an even positive integer
(c) a rational number other than a positive integer
(d) an irrational number
[2012]
22. The term independent of x in the expansion of
x +1
x - 1 ˆ 10
Ê
ÁË x 2 3 - x1 3 + 1 x - x1 2 ˜¯ is
(a) 120
(b) 210
(c) 310
(d) 4
[2013]
23. If the 7th term in the binomial expansion of
9
ˆ
Ê 3
ÁË 1 3 + 3 ln x˜¯ , x > 0 is equal to 729, then x can be:
84
26. The sum of the rational terms in the binomial expansion of (21/2 + 31/5)10 is:
(a) 25
(b) 32
(c) 9
(d) 41
[2013, online]
27. If the coefficients of x3 and x4 in the expansion of
(1 + ax + bx2) (1 – 2x)18 in powers of x are both zero,
then (a, b) is equal to:
251ˆ
272 ˆ
Ê
Ê
(b) Ë 14,
(a) Ë 14,
¯
3
3 ¯
272 ˆ
251ˆ
Ê
Ê
(c) Ë 16,
(d) Ë 16,
[2014]
3 ¯
3 ¯
28. The number of terms in the expansion of
(1 + x)101 (1 + x2 – x)100 in powers of x is
(a) 302
(b) 301
(c) 202
(d) 101
[2014, online]
29. The coefficients of x 50 in the
(1 + x)1000 + x(1 + x)999 + x2(1
(1000 )!
(b)
(a)
(50 )! (950 )!
(c)
(1001)!
(d)
(51)! (950 )!
binomial expansion of
+ x)998 + . . . + x1000 is:
(1000 )!
( 49)! (951)!
(1001)!
(50 )! (951)!
[2014, online]
55
xˆ
Ê
30. If Ë 2 + ¯ is expanded in the ascending powers of x
3
and the coefficients of powers of x in two consecutive
terms of the expansion are equal, then these terms are:
(b) 8th and 9th
(a) 7th and 8th
th
th
(c) 28 and 29
(d) 27th and 28th
[2014, online]
5
31. If 1 + x4 + x5 = Â ai (1 + x)i, for all x in R, then a2
i=0
is:
(a) – 4
(c) – 8
(b) 6
(d) 10
[2014, online]
Mathematical Induction and Binomial Theorem
32. The coefficient of x1012 in the expansion of (1 + xn +
x253)10, (where n £ 22 is a positive integer), is
(a) 1
(b) 10C4
(c) 4n
(d) 253C4 [2014, online]
x in the
binomial expansion of (1 - 2 x )
1 50
(3 + 1)
2
1 50
(3 – 1)
(c)
2
(a)
50
(a) 64
(c) 243
37. For x
(1 + x)
1 50
(3 )
2
1 50
(d)
(2 + 1)
2
+ x(1 + x)2015 + x2 (1 + x)2014 + … + x2016
i= 0
(a)
2017 !
17 !2000!
(b)
2016!
17 !1999!
(c)
2016!
16!
(d)
2017 !
2000!
[2015]
[2016 online]
x–2 and x–4 in the expansion of
18
[2015 online]
35. The term independent of x in the binomial expansion
8
1ˆ
Ê 1
ˆÊ
of Á1 - + 3x5 ˜ Á 2 x 2 - ˜ is:
Ë
¯Ë
x
x¯
(a) 400
(b) 496
(c) – 400
(d) – 496
–1, if
i
= Â ai x , then a17 is equal to:
(b)
(b) 7th
(d) 9th
R, x
2016
[2016]
2016
is:
binomial expansion of (1 + x)n are in the ratio 1 : 7 :
(a) 6th
(c) 8th
(b) 2187
(d) 729
7.35
[2015 online]
36. If the number of terms in the expansion of
Ê 1
ˆ
Á x 3 + 1 ˜ , (x > 0), are m and n
1˜
Á
Ë
2x3 ¯
m
is equal to:
n
(a) 27
(b) 182
5
4
(d)
(c)
4
5
[2016, online]
n
Ê 2 4ˆ
ÁË1 - + 2 ˜¯ , x
x x
0, is 28, then the sum of the
Previous Years' B-Architecture Entrance
Examination Questions
1 ˆn
Ê
1. If in the expansion of Á x 3 - 2 ˜ , the sum of the
Ë
x ¯
5
10
coefficients of x and x is 0, then the coefficient of
the third term is:
(a) 455
(b) 105
(c) 605
(d) 120
[2006]
(a) 1
(c) 7
5. The value of the sum
8
1
Ê 8ˆ
 ( j + 1) ( j + 2) ÁË j˜¯
j=0
2. If (1 + x) (1 + x + x2) (1 + x + x2 + x3) . . . (1 + x
+ x 2 + . . . + x n) = a 0 + a 1 x + a 2 x 2 + . . . + a m x m,
then the value of a1 is
(a) m + 1
(b) (n + 1)
(c) n
(d) m
[2006]
1003
1013
(b)
90
90
1023
1033
(c)
(d)
90
90
6. Statement-1: The coefficients of xn in
3. If the ratio of the 7th term from the beginning to the
7th term from the end in the expansion of (21/3 +
3–1/3)n is 1/6, then the value of n is
(a) 9
(b) 12
(c) 6
(d) 3
[2007]
4. The remainder when 7128 is divided by 10 is
(b) 3
(d) 9
[2008]
is
(a)
n
 (-1) j
(2x + 5)n – j (5 – 2x)j is (n + 1) 2n
j=0
Statement-2:
8
Â
j=0
an- j b j =
a n +1 - b n +1
,aπb
a-b
[2008]
7.36
Complete Mathematics—JEE Main
and coefficient of (a x + b )m is a m
[2009]
n
2
7. If x = a0 + a1 (1 + x) + a2 (1 + x) + . . . + an
(1 + x)n
= b0 + b1 (1 – x) + b2 (1 – x)2 + . . . + bn (1 – x)n
then for n = 101, (a50, b50) equals:
(b) (101C50, – 101C50)
(a) (– 101C50, 101C50)
(d) (101C50, 101C50) [2009]
(c) (– 101C50, – 101C50)
È1
8. If the third term in the expansion of Í + x
Îx
x > 1, is 1000, then x equals
log10 x ˘
5
˙˚ ,
(a) 10
(b) 1
1
(d) 100
[2010]
(c)
10
9. If the sum of the coefficients in the expansion of (x +
y)n is 2048, then the greatest coefficient in the expansion is:
(b) 11C6
(a) 10C6
11
(c) C7
(d) 12C6
[2011]
10. If ai(i = 0, 1, 2, . . ., 16) be real constants such that
for every real value of x, (1 + x + x2)8 = a0 + a1x +
a2x2 + . . . + a16x16, then a5 is equal to:
(a) 502
(b) 504
(c) 506
(d) 508
[2012]
24
2 12
12
24
11. The coefficient of t in (1 + t ) (1 + t ) (1 + t )
is:
(a) 12C6 + 13
(b) 12C6 + 2
(d) 12C6
[2013]
(c) 12C6 + 1
12. Sum of the last 30 coefficients of powers of x in the
binomial expansion of (1 + x)59 is:
(b) 228
(a) 229
59
29
(c) 2 – 2
(d) 258
[2014]
m
13. If in the binomial expansion of (1 – x) (1 + x)n, the
x and x2
the ratio m : n is equal to:
(a) 10 : 7
(b) 8 : 11
(c) 10 : 13
(d) 7 : 10
[2015]
n, if the mean of the binomial
a + b)2n – 3 is 16,
then n is equal to:
(a) 5
(b) 7
(c) 9
(d) 4
[2016]
Answers
Level 1
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
63.
67.
(c)
(b)
(b)
(b)
(b)
(d)
(a)
(c)
(a)
(c)
(b)
(a)
(a)
(b)
(a)
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
64.
68.
(a)
(b)
(b)
(a)
(a)
(a)
(a)
(c)
(b)
(c)
(b)
(b)
(d)
(c)
(d)
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
57.
61.
65.
69.
(b)
(d)
(b)
(d)
(a)
(a)
(a)
(b)
(a)
(b)
(d)
(b)
(d)
(a)
(a)
1. (a)
2. (b)
3. (a)
4. (c)
5. (c)
6. (d)
7. (b)
8. (c)
9. (a)
10. (d)
(a)
(a)
(a)
(a)
(c)
(c)
(b)
(a)
(d)
(b)
(a)
(d)
(d)
(d)
(d)
Level 2
71. (d)
75. (b)
79. (a)
72. (a)
76. (b)
80. (a)
73. (a)
77. (c)
74. (a)
78. (d)
Previous Years' AIEEE/JEE Main Questions
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
(b)
(a)
(b)
(a)
(d)
(d)
(d)
(d)
(c)
(a)
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
(a)
(b)
(a)
(b)
(c)
(b)
(d)
(b)
(b)
(b)
3.
7.
11.
15.
19.
23.
27.
31.
35.
(b)
(b)
(b)
(d)
(d)
(b)
(c)
(a)
(a)
4.
8.
12.
16.
20.
24.
28.
32.
36.
(c)
(a)
(a)
(a)
(a)
(a)
(c)
(b)
(b)
Previous Years' B-Architecture Entrance
Examination Questions
1.
5.
9.
13.
(d)
(b)
(b)
(d)
2.
6.
10.
14.
(c)
(a)
(b)
(a)
3. (a)
7. (a)
11. (b)
Hints and Solutions
Concept-based
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
62.
66.
70.
Concept-based
Ê kˆ
1. Tr + 1 = 5Cr (x2)5 – r Ë ¯
x
r
= 5Cr kr x10–3r
4. (a)
8. (d)
12. (d)
Mathematical Induction and Binomial Theorem
set 10 – 3r = 1 or r = 3. Therefore,
fi (n – 14) (n – 23) = 0
5
Thus, n can be 23
3
3
C3 k = 270 fi k = 27 fi k = 3.
3
5
Ê 1ˆ
2. T4 = nC3(px)n – 3 Ë ¯ = nC3 pn – 3 xn – 6 =
2
x
5
¤ n – 6 = 0 and 6C3 p3 =
¤ n = 6, p = 1/2
2
3. 20C3r + 3 = 20C2r – 3 fi (3r + 3) + (2r – 3) = 20 fi r = 4
9. (p + 2)th term from the end
= [2n + 2 – (p + 1)]th = (2n – p + 1)th
from the beginning
2n - p
Ê 2ˆ
and T2n – p + 1 = 2n + 1C2n – p (x2)2n + 1 – (2n – p) Á - 2 ˜
Ë x ¯
= 2n + 1C2n – p (– 2)2n – p x2p + 1 – 2n
4. Sum of the coefficients
= (1 + 2)n = 3n = 38 fi n = 8.
T4 = 8C3(2x)3 = 8C3 23 x3
10.
Thus, coefficient of fourth term is 448.
1 ˆ5
Ê
5. (1 + x2)10 Á x 2 + 2 + 2 ˜
Ë
x ¯
2 10
= (1 + x )
Ê
Ëx +
15
n (n - 1) 2
a
2
7. Tr + 1 = (r + 1)th term in (1 + x2)10 =
10
\ coefficient of x10 in (1 + x2)10 is
C 5.
Cr (x2)r
2 ˆ 10
Ê
Also, tr + 1 = (r + 1)th term in Ë x + ¯
x
=
=
10
Cr (x)
r
10 – r
Cr 2 x
Ê 2ˆ
Ë x¯
r
11. As there is just one middle term, n must be even and
Ên ˆ
middle term is Ë + 1¯ th term, and it is given by
2
Ê 1ˆ
Tn/2 + 1 = nCn/2 (x2)n – n/2 Ë ¯
x
12. We have
20
Cr
10 – 2r
\ term independent of x is
Required ratio is
10
C5 :
10
10
C5 (25)
5
C5(2 ) = 1 : 32
8. 2(nC9) = nC8 + nC10
n-9
9
+
n-8
10
fi 20(n – 8) = 90 + (n – 8) (n – 9)
fi n2 – 37n + 322 = 0
n2
= nCn/2 x n/2
Now, nCn/2 xn/2 = 924x6
fi n/2 = 6 fi n = 12
[Also, note that 12C6 = 924]
Tr + 1 =
For this term to be independent of x, set 10 – 2r = 0
or r = 5.
fi2=
= 1 (15) (16) = 120
2
Level 1
4 n -1
80 n (n - 1) a 2
=
fi =
fin=5
2
2
5
n
100
n a
and a = 2
10
15
Ê C ˆ
(16 - r )
=
Â
˜
r -1 ¯
r =1
 r ÁË C r
r =1
= coefficient of x15 in (1 + x2)15 = 0.
10
16 - r
r
fiS=
1 ˆ5
Ê
\ coefficient of x10 in (1 + x2) Á x 2 + 2 + 2 ˜
Ë
x ¯
6. 10 = na, 40 =
Cr
15!
(r - 1)! (16 - r )!
=
¥
Cr -1
r ! (15 - r )!
15!
=
1ˆ 5 1
= 5 (1 + x2)15
x¯
x
7.37
5
r
( 5)
=
20
Cr
1
( 5 )r - 1
Suppose (r + 1)th term is the greatest term. Then
Tr + 1
Tr +1
> 1 and
>1
Tr
Tr + 2
fi
20 - r 1
r+2
5 >1
> 1 and
r +1 5
19 - r
fi 20 – r >
5 r + 5 and r 5 + 2 5 > 19 – r
fi 19 – 2 5 <
(
5 + 1) r < 20 –
5
1
1
(19 – 2 5 ) ( 5 – 1) < r < (20 – 5 ) ( 5 – 1)
4
4
1
1
fi
( 21 5 – 29) < r <
( 21 5 – 25)
4
4
fi
Complete Mathematics—JEE Main
7.38
fi y = x–3
fi 4.48 < r < 5.49 fi r = 5,
4
Ê 1 ˆ
C5 Á
=
˜
Ë 5¯
13. S(q ) = ( 2 – sinq + cos q)2n
Maximum possible value of
20
\ greatest term =
2 – sin q + cos q =
p
cos q)
sin
4
2 –
=
2 –
20
Ê 1ˆ
C5 Ë ¯
25
2 (cos
p
sin q –
4
14. Note that Tr = nCr xn – r ar.
2
n
(ia) + C3 x
n
+ . . . + Cn (ia)
For (r + 1)th term to be independent of x, set 12 – 6r
= 0 or r = 2
= 4C2 52 = (6) (25) = 150
18. (1 + 5x + 10x2 + 10x3 + 5x4 + x5)20
= ((1 + x)5)20 = (1 + x)100
(x + ai)n = nC0 xn + nCr xn – 1 (ia)
+ C2 x
= 4Cr 5r x12 – 3r – 3r
\ required constant term
is 2 + 2 = 2 2
\ Maximum possible value of S(q ) is (23/2)2n = 8n
n–2
(x3 + 5x –3)4 is
Tr + 1 = 4Cr (x3)4 – r (5x – 3)r
2 sin (q – p /4)
n
(r + 1)th term in the expansion of
m = number of terms = 101.
n–3
(ia)
3
n
We have
2m = 2101 = (25)20 (2)
= (T0 – T2 + T4 – . . . )
= (30 + 2)20 (2) = (10 m1 + 220) (2)
+ i(T1 – T3 + T5 – . . .)
Let k = Unit's digit of 2m is
fi (x2 + a2)n = |(x + ai)n|2
= units digit of 221
= (T0 – T2 + T4 – . . .)2 + (T1 – T3 + T5 – . . .)2
Now, 221 = (25)4 (2) = (30 + 2)4 (2)
\ k = unit’s digit of 25 = 2.
15. For 0 £ r < 2n,
(1 + x)4n = (1 + x)2n + r (1 + x)2n – r
2
19. Suppose
= (A0 + A1x + A2x + . . . + A2n + r x
2
(B0 + B1 x + B2 x + . . . + B2n – r x
where Ak =
and Bk =
2n + r
2n – r
2n + r
2n – r
)
and (1 + x)n =
k =0
Bn – 1 + . . . + Ar B2n – r
Now,
= coefficient of x2n on LHS
=
fi
Br + 1
Ar
C2n.
Thus, An Bn <
4n
<
4n
C2n
2 17
16. Sum of the coefficients = (2 – 3c + c )
2 17
We are given (2 – 3c + c )
fi log3 y = –
3
fi
( x3 2 )
log3 ( x 3 2 )
log3 3
Thus,
=0
fi c2 – 3c + 2 = 0 fi c = 1, 2
17. Let y = 3- log
=
=
C2n
( 2 n + r Cn ) ( 2 n - r Cn )
 Ck x k , where Ck = nCk.
We have
Pn +1
ÊB ˆ
= B0 Á 1 ˜
Ë A0 ¯
Pn
A2n B0 + A2n – 1 B1 + . . . + An Bn
4n
k =0
n
Ck (0 £ k £ 2n + r)
Ck (0 £ k £ 2n – r)
+1
 Bk x k , where Bk = n + 1Ck
)
Coefficient of x2n on the RHS
+ An
n +1
(1 + x)n + 1 =
Pn + 1
Pn
=
Ê B2 ˆ Ê Bn + 1 ˆ
ÁË A ˜¯ . . . ÁË A ˜¯
1
n
r ! (n - r )!
(n + 1)!
◊
n!
(r + 1)! (n - r )!
n +1
r +1
(0 £ r £ n)
(n + 1)n
r!
2120 ( n + 1)n
fi n = 20
=
20!
r!
20. Coefficient of the term independent of x in the expan3/2
= – 2log3 (x )
Ê x ˆ
sion of (1 + x)2n Á
Ë 1 - x ˜¯
-2 n
Mathematical Induction and Binomial Theorem
32. Positive terms are
(1 + x )2 n (1 - x )2 n
or
4n
C0, 4nC4x4, 4nC8x8, º, 4nC4n x4n
\ number of positive terms = (n + 1).
x2n
= coefficient of x2n in (1 – x2)2n
= coefficient of t n in (1 – t)2n
=
2n
Cn(–1)n
10 – r
21. Tr + 1 =
10
x
Cr Ê ˆ
Ë2 ¯
r
Ê 3ˆ =
–
ÁË 2 ˜¯
x
10
Cr
( –3)r
10 – r
2
For the coefficient of x set 10 – 3r = 4
fi r = 2.
x10 – 3r
18
C2r + 3 = 18Cr – 3
fi 2 r + 3 = r – 3 or 18 = (2r + 3) + (r – 3)
fi
r = – 6 or r = 6
24. (x + a)
200
+ (x – a)
200
C0 x200 +
= 2[
200
C200 a200]
These are 101 terms.
Cr
fi
+ 1
3 r – 1 + r + 1 = 2n
1
n(n – 1)p2x2 + º
2
27. (1 + px)n = 1 + npx +
28.
È
+
10
10
2
1
3
= Ê1 – ˆ + Ê1 – ˆ
Ë 2¯
Ë 4¯
29. Given expression
=
C1( 2 ) 2n +
2n + 1
C3 ( 2 ) 2n
=
50
50
C3(100)47 + º +
50
C49(100)]
+ a positive term.
fi
a < b.
38. Clearly A0 = 1.
210 + 1
220
Putting x
= 1 and x = – 1, we get
= A0 + A1 + A2 + º + A12
0
6
and (– 2) = A0 – A1 + A2 + º + A12
Adding we get
64 = 2[A0 + A2 + º + A12]
fi A2 + A4 + º + A12 = 31
39. Put x = 1.
30. Use (x + a)6 + (x – a)6
= 2[x6 + 6C2 x4a2 + 6C4 x2 a4 + 6C6 a6]
2
+
= (100 + 1)50 – (100 – 1)50
fi b = a + (a positive term)
Ck
Putting a =
of 9.
– 2
º+
C2n + 1] is an integer
fi [R] + f – F = 2m fi f – F = 2m – [R] is an
integer.
But – 1 < f – F < 1. Thus, f – F = 0
\ Rf = RF = 1
36. The coefficient of x60 in (1 + x)51 (1 – x + x 2)50
= the coefficient of x60 in (1 + x) (1 – x3) 50
= 100
3 ˘
˙
22r ˚
= number of ways of choosing k persons out m men
and n women
m+n
+ 1
= 2[50C1(100)49 +
r
 (–1)r ( 10 Cr ) ÍÎ 1r
r=0
x = 100.
35. Let F = ( 2 – 1)
. Note that 0 < F < 1.
Also, R – F = 2m, where
= 50C20
37. 10150 – 9950
1
n(n – 1) p2 = 264
2
Solve to obtain n = 12, p = 2.
\ np = 24,
10
or
10
2n + 1
\ Coefficient of x5 is (5)(4) + (4)(10) = 60
2n
1
fi x=
m = 2[2n
4
= (1 + 5x2 + 10x4 + º) (1 + 4x + 6x2 + 4x3 + º)
26. 2nC3r – 1 =
fi n = 2r.
10
2n + 1
C2 x198 a2 + º +
25. (1 + x ) (1 + x)
( x log x )2 = 1000
fi (2t – 3)t = 2 where t = log10x
fi 2t2 – 3t – 2 = 0
fi (2t + 1) (t – 2) = 0
200
200
2 5
5–2
2 log x - 3
fi 10 x 10
= 1000
n
22. a = Cr and b = Cn – r
\ a=b
23.
1 n
1 2n
33. ÊÁ x 2 – 2 + 2 ˆ˜ = Ê x – ˆ
Ë
Ë
x¯
x ¯
Number of terms in the expansion is (2n + 1) out of
which just one is independent of x.
34. T3 = 5 C2 Ê 1 ˆ
Ë x¯
4
n
7.39
3
x – 1 , we find it to be a polynomial
3 7
3 7
31. (1 + 3x + 3x + x ) = ((1 + x) ) = (1 + x)
21
40. Let
R=
Put
F=
(
(
6
3 + 1) and R – [R] = f.
6
3 – 1) , then 0 < F < 1 and
7.40
Complete Mathematics—JEE Main
R+F=
41.
42.
43.
44.
45.
6
(
6
n
3 + 1) + ( 3 – 1)
fi
= 2[33 + 6C2 (32) + 6C4(3) + 1]
= 2[27 + 135 + 45 + 1] = 416
fi f + F = 416 – [R] is an integer.
Now, show that f + F = 1.
Use (x2 + 6x + 9)30 = (x + 3)60
See Theory.
Number of terms in (x + y + z)n
= number of non-negative integral solution
of p + q + r = n
1
= n + 2C2 = (n + 2) (n + 1)
2
Similar to Question 43.
Tr + 1 , the (r + 1)th term in the expansion
11
Ê 2 1 ˆ is
Ë ax + bx ¯
1 r
Tr + 1 = 11Cr(ax2)11 – r Ê ˆ =
Ë bx ¯
11
a11 – r
Cr
br
x 22 – 3r
1
\ Coefficient of x 7 in Ê ax 2 + ˆ
Ë
bx ¯
is
11
C5
a
tr + 1 =
\
2n = 128 = 27 fi n = 7
Now, Tr + 1, the (r + 1)th term in the expansion of
1ˆ 7
Ê
is
x
+
Ë
x¯
r
1
Tr + 1 = 7Cr x7 – r Ê ˆ = 7Cr x7 – 2r
Ë x¯
For the coefficient of x, we set 7 – 2r = 1 fi r = 3.
1 7
Thus, coefficient of x in the expansion of Ê x + ˆ
Ë
x¯
is 7C3 = 35.
n
48.
C4 an – 4 (– b)4 + nC5 a n – 5 (– b)5 = 0
fi nC 4 a n – 4 b 4 = nC 5 a n – 5 b 5
fi
11
1 r
Cr ( ax )11 – r ÊÁ – 2 ˆ˜
Ë bx ¯
We are given
50.
a11 – r
11
C5
a6
b5
=
11
C6
a5
2
b6
Differentiating both the sides with respect to t, we get
n(a + t)
=
 r(
n
Cr ) a
r =1
t
 r ( n Cr ) x n – r yr
r=0
n
Cr
= 2 Â ( n – r + 1) = n ( n + 1)
Cr – 1
r =1
C
\ ÊÁ 1 + 1 ˆ˜
Ë
C0 ¯
r=0
n(x + y)n – 1 y =
1 11
(3 – 1)
11
51. From the previous question, we have
Cr ˆ
n +1
Ê
ÁË 1 + C ˜¯ = r
r –1
n –r r –1
Multiplying both the sides by t and putting a = x
and t = y, we get
n
dx =
Cr
n!
(r – 1)! (n – r + 1)!
=
.
Cr – 1 r ! (n – r )!
n!
fi 2Â r
ab = 1.
r=0
n
C4
a n–4
=
b
5
n – r +1
r
C
fi r r = n–r+1
Cr – 1
b6
 n Cr a n – r t r
n–1
fi
=
a5
n
46. We have (a + t)n =
10
Ú0 (1 + x )
n
fi
C5
2
=
\ Coefficient of x –7 in the expansion of = 11 C6
b
n
Ú0 (C0 + C1 x + C2 x 2 + C10 x10 ) dx
is
Cr
n–5 5
n
=
49. Given expression is equal to
(–1)r x11 – 3r
br
For the coefficient of x –7, set 11 – 3r = – 7 fi r = 6.
=
a n – 4 b4
a
.
11
11
x + y = 1]
1ˆ n
Ê
is equal to 2 n.
x
+
Ë
x¯
Next, tr + 1 , the (r + 1)th term in the expansion of
1 ˆ
Ê
ÁË ax – 2 ˜¯
bx
[
47. Sum of the coefficients in the expansion of
6
b5
= ny
r=0
For the coefficient of x7, set 22 – 3r = 7 fi r = 5.
11
 r ( n Cr ) x n – r yr
52. C0 –
=
1
Cn ˆ
(n + 1)n
C2 ˆ Ê
Ê
+
1
=
1
+
ÁË
Cn – 1 ˜¯
n!
C1 ˜¯ ÁË
1
1
C1 + C2 – 3
5
Ú0 ÈÎC0 – C1 x 2 + C2 x 4 – ˘˚ dx
=
1
Ú0 (1 – x 2 )
n
dx
Mathematical Induction and Binomial Theorem
53. (1 + 2x 2 + x 4) (1 + nC1x + n C 2x2 + n C 3 x 3 + º)
2
3
= a 0 + a 1x + a 2x + a 3x + º
fi a1 = nC1, a2 = n C 2 + 2, a3 = nC3 + 2(nC1 )
Now, a1, a2, a3 are in A.P. implies
\ (x 2 – a 2)n = (x + a)n (x – a)n = (A + B) (A – B)
= A 2 – B 2.
n = 2, 3, 4
 ( 4 n + 1Ck + 4 n + 1C2 n – k )
2n
(4n + 1Ck) +
Â
4n + 1
4n + 1
C2n
55. Coefficient of t
24
[See Theory]
2 12
in (1 + t )
= Coefficient of t 24 in (1 +
12
12
24
(1 + t ) (1 + t )
C1 t 2 +
12
C2 t 4 + º)
(1 + t 12 + t 24)
12
=
12
C12 +
12
C6 + 1 =
C6 + 2
56. T3 = 5 C2 x 5 – 2 ( x log10 x )2 = 106 fi x 2 log10 x + 3 = 105
fi t (2t + 3) = 5 where t = log10 x fi t = – 5/2, 1
57. a =
(10 + 1) (2 / 3) 22
1+ 2/3
=
5
fi
[a] = 4
Thus, the greatest term is 5th term and it is given by
10
2
C4 Ê ˆ
Ë 3¯
58. a =
4
2
= 210 Ê ˆ
Ë 3¯
4
=
11
Ê 2ˆ
Cr Á x ˜
Ë a¯
Cr(–1)r ar
r
Ê bˆ
–
Ë x¯
– 11 r 22 – 3r
r = 6 and r = 5
C6(– 1)6 a –5 b6 +
1ˆ8
x¯
Using this show that 3(52k + 3) + 23k + 4 is divisible by
17.
64. tr = r (r + 1) (r + 2) = (r + 1) {(r + 1)2 – 1}
= (r + 1)3 – (r + 1)
 tr =
n +1
 r3 –
r =1
n +1
Âr
r =1
=
1
1
(n + 1)2 (n + 2 )2 – (n + 2 ) (n + 1)
4
2
=
1
(n + 1) (n + 2 ) (n2 + 3n + 2 – 2 )
4
=
1
n ( n + 1) ( n + 2 ) ( n + 3) .
4
bx
We are given
11
1ˆ8
x¯
Assume 3(52k + 1) + 23k + 1 = 17m for some m.
r =1
For coefficient of x 4 and x 7, we put 22 – 3r = 4
and 22 – 3r = 7
fi
1ˆ 4 Ê
x–
x¯ Ë
63. For n = 1, 3(52n + 1) + 23n + 1 = 3(53) + 24 = (17) (23)
n
11 – r
1ˆ 4 Ê
x–
x¯ Ë
= ( 4C 0 ) ( 8C 8) – ( 4C 1 ) ( 8C 6 ) + ( 4C 2) ( 8C 4) – ( 4C 3) ( 8C 2)
+ (4C4) (8C0) = 198
fi
=4
3 + 5 (1 / 5)
Thus, 4th and 5th terms are the greatest, and T5 =
15
C4(3)11
11
1 ˆ 12 Ê
= x+
Ë
x¯
1
¥ ÊÁ 8 C0 x8 – 8C1 x 6 + 8C2 x 4 – + 8C8 8 ˆ˜
Ë
x ¯
\ Coefficient of the term independent of x
(15 + 1) (5) (1 / 5)
59. Tr + 1 =
1ˆ 4 Ê
x–
x¯ Ë
1
1
= ÊÁ 4 C0 x8 – 4C1 x 4 + 4C2 – 4C3 4 + 4C4 8 ˆ˜
Ë
x
x ¯
C2n
k=0
= 2 4n +
62. Ê x +
Ë
1 4
= ÊÁ x 2 – 2 ˆ˜ Ê x –
Ë
x ¯ Ë
k=0
=
fi ab = 1
a 5 a6
60. Numerator = 193 + 63 + 3(19) (6) (19 + 6)
= (19 + 6)3 = 253
61. We have (x + a)n = A + B and (x – a)n = A – B.
n
54.
b5
Thus, given expression = 1.
fi n3 – 9n2 + 26n – 24 = 0
fi
=
and denominator = (3 + 2)6 = 56
2( n C2 + 2) = n C1 + nC3 + 2( n C1)
1
fi n(n – 1) + 4 = 3n + n(n – 1) (n – 2)
6
fi 6(n2 – n) + 24 = 18n + n3 – 3n2 + 2n
fi (n – 2) (n – 3) (n – 4) = 0
b6
fi
7.41
11
C5 (– 1)5 a– 6 b5 = 0
Remark
We can also prove this by using the principle of mathematical induction.
65. Use a2m + 1 + b2m + 1
= (a + b) (a2m – a2m – 1 b + a2m – 2 b2 – º
– ab 2m – 1 + b2m]
Complete Mathematics—JEE Main
7.42
n
Â
66. Let S =
Cr sin (rx) cos [(n – r)x]
(1)
r=0
But for 1 £ r £ p – 1, neither r ! nor (p – r)! is divisible by p.
Using Cr = Cn – r , we can write
n
Â
S=
Ê pˆ
As p | p!, we get p |r! (p – r)! Á ˜
Ë r¯
Cn – r sin (rx) cos [(n – r)x]
Ê pˆ
pÁ ˜
Ë r¯
\
r=0
n
Â
=
Cr sin [(n – r)x] cos (rx)
(2)
r=0
1
5–2=
We have
5+2
Adding (1) and (2), we get
fi0< 5 –2<1
n
Â
2S =
Cr {sin (rx) cos [(n – r)x] +
(
Let 2 + 5
r=0
sin [(n – r)x] cos (rx)}
Â
Cr sin (nx) = sin (nx)
Â
Cr
= 2 sin (nx)
S = 2n – 1 sin (nx)
Ê
Á1 +
x Ë
n
xˆ
˜
3¯
=3
2n
È
Ê
n
ÍCoefficient of x in the expansion of ÁË 1 +
Î
x ˆ 2n ˘
˜ ˙
3¯ ˚
n
3n (2n )!
Ê 2 nˆ Ê 1 ˆ
= 32n Á ˜ Á ˜ =
Ë n ¯ Ë 3¯
n! n!
(
5+2
n
) –(
5–2
Ê nˆ Ê nˆ
Since n is an odd prime, each of Á ˜ , Á ˜ , º
Ë 1 ¯ Ë 3¯
n
Ê
ˆ
ÁË n – 2˜¯ is divisible by n.
Thus RHS of (1) is divisible by 20 n
Also, F – f is an integer. Since
0 < F < 1 and 0 < f < 1, we get – 1 < F – f < 1.
As F – f is an integer, we get F – f = 0 or F = f.
(
)
n
– 2n
+ 1
is N – 2n
+ º + (1 + x) }
n
= coefficient of x m in
69. We have
Ê pˆ
p!
ÁË r ˜¯ =
r ! ( p - r )!
We can write
Let k =
(
3+ 2
}
=
n+1
in [(1 + x)
m
– (1 + x) ]
51 3
(
50
)
Ê
3ˆ
= 225 Á 1 +
2 ˜¯
Ë
50
.
(50 + 1) 3 / 2
–1
1+ x –1
= coefficient of x
Ê n + 1ˆ
= Á
Ë m + 1˜¯
,
70. That Statement-2 is false can be seen from theory.
[using Statement-2]
m+1
+ 1
which is divisible by 20 n.
= coefficient of x m in {(1 + x)m + (1 + x)m + 1
(1 + x)m
n
)
n is odd]
68. LHS of (1)
n – m +1
< 1.
(1)
\ integral part of 2 + 5
[using Statement-2]
{(1 + x )
n
[
n
9
Ê
ˆ
\ Coefficient of the term independent of x Á x + + 6˜
Ë
¯
x
2n
)
˘
Ê n ˆ
++ Á
( 5) 2 n – 1 ˙
˜
Ë n – 2¯
˚
n
3
5–2
( )
n
Ê x2 + 6 x + 9ˆ
9
(3 + x )2 n
Ê
ˆ
67. Á x + + 6˜ = Á
=
˜¯
Ë
¯
x
x
Ë
xn
=
(
ÈÊ nˆ
Ê nˆ
= 2 ÍÁ ˜ 5(n – 1) / 2 (2 ) + Á ˜ 5(n – 3) / 2 23
Ë
¯
Ë 3¯
1
Î
r=0
n
2n
0<f=
= N + F where 0 < F < 1. Now,
n
r=0
fi
n
N + F – f – 2n + 1 =
n
=
)
fi
1+ 3/2
(
3– 2
)
)(
3– 2
51 (0.5505)
28.07
3+ 2
)
(
= 51 3 – 6
)
fi [k] = 28.
fi
p
r!(p – r)! ÊÁ ˆ˜ = p!
Ë r¯
Thus, the greatest term in the expansion of
is the 29th term and it equals
(
3+ 2
50
)
Mathematical Induction and Binomial Theorem
25 Ê 50ˆ Ê
3ˆ
2 Á ˜Á
Ë 28¯ Ë 2 ˜¯
28
+ ( nC n – 2) ( n + 1C 3) +
+ (nC0) (n + 1 Cn + 1)]
1
1
(2n + 1 Cn + 1 ) =
(2n + 1Cn)
=
n +1
n +1
Ê 50ˆ 14 11
= ÁË 22˜¯ 3 2
Level 2
n
71. Note that x 2 - 1 is defined if x2 ≥ 1 i.e. if x £ – 1
or x ≥ 1 But cos–1 x is defined for |x| £ 1.
76.
n
n
n
 (n - k )(k + 1) = n  (k + 1) -
=
k=0
k=0
n
 k2 -
k =1
Âk
k =1
Ê 1ˆ
Ê 1ˆ
= n Á ˜ (n + 1) (n + 2) – Á ˜ n (n + 1) (2n + 1)
Ë 2¯
Ë 6¯
Ê 1ˆ
– Á ˜ n (n + 1)
Ë 2¯
1
n (n + 1) (n + 2)
6
=
78. Put x – 3 = t, so that
p–2
2
p
= C1 a
b + C2 a
b +
+ Cp–1 ab
For 1 £ r £ p – 1,
(pCr) (r!) ( p – r)! = p!
As p is prime, p | p! but
p (r!) and p (p – r)!
\ p | pC r
Thus, p divides (a + b) p – ap – b p.
75. We have
p–1
1
1
1
C2
Thus, C02 + C12 + C12 +
+
n +1 n
2
3
1
1
1
= C0 ÊÁ C0 ˆ˜ + C1 ÊÁ C1 ˆ˜ + C2 ÊÁ C2 ˆ˜
Ë1 ¯
Ë2 ¯
Ë3 ¯
Ê 1
ˆ
C
+ Cn Á
Ë n + 1 n ˜¯
1
=
[(nCn) (n + 1C1) + (nCn – 1) (n + 1C2)
n +1
2n
2n
 br t r
=
 ar (1 + t )r
r=0
r=0
fi bn = coefficient of tn of the R.H.S.
= an (nCn) + (an + 1) (n + 1Cn) +
+ (a2n) (2nCn)
n
 ak (n + k Ck )
=
=n+1
k=0
79. (1 + x)41 (1 – x + x2)40
= (1 + x) [(1 + x) (1 – x + x2)]40
(n + 1) (nCr) = (r + 1) ( n + 1Cr +1)
1
1
fi
( nC r) =
( n + 1C r + 1)
r +1
n +1
+
Ê n + 0ˆ
+ 12 Á
Ë 1 ˜¯
Ê 2 ˆ
Ê 1ˆ
+
S = n2 Á ˜ + (n - 1)2 Á
Ë n¯
Ë n - 1˜¯
50
C0 1050 – 50C1 1049 +
C47 103
+ 50C48 102 – 50C49 (10) + 50C50
= 1000m + 122000 + 1 = 1000 (m + 122) + 1
where m is a natural number
Thus, hundred’s digit of 3100 is 0.
p
2
 ( n Ck )
Ck
k +1
=
,
Ck + 1
n-k
77. As
50
p–1
((n - k )!)2
2
k=0
73. 3100 = 950 = (10 – 1)50
p
2
(2 n) ! n Ê
n!
ˆ
=
Â
Á
Ë
n ! n! k = 0 k ! (n - k ) !˜¯
= (2nCn) (2nCn) = (2nCn)2
+ 40C40 x40 ] ¥ [40C0 + 40C1 x +
+ 40C40 x40]
= Coefficient of x25 in (1 – x)40 (1 + x)40
= Coefficient of x25 in (1 – x2)40 = 0
74. (a + b)p – ap – bp
Â
k = 0 ( k !)
n
72. ( 40C0) ( 40C15) – ( 40C1) ( 40C16) + ( 40C2) ( 40C17)
+
– (40C25) (40C40)
= (40C0) ( 40C25 ) – (40C1) (40C24) + (40C2) (40C23)
– (40C25) (40C0)
+
= Coefficient of x25 in [40C0 – 40C1 x + 40C2 x2 –
(2 n) !
= (2nCn)
Thus, only possible values of x are 1 and – 1. But
these do not satisfy the given equation.
=
7.43
= (1 + x) (1 – x3)40
Coefficient of x50 in (1 + x)41 (1 – x + x2)40
= Coefficient of x50 in (1 – x3)40
+ Coefficient of x49 in (1 – x3) = 0 + 0 = 0
80.
 (n Ci )
( n + 1C j)
i< j
n
=
 (n Ci )
[ n + 1C i
+1
+
n+1
Ci
+2
i=0
n+1
C n + 1]
+
+
Complete Mathematics—JEE Main
7.44
tr + 1 =
n
=
Â(
n
n
n
Ci ) [( Ci + Ci
+ 1)
i=0
2
 ( n Ci )
+ 2Â
i= 0
n
n
+ ( C i + 1 + C i + 2)
n
+
n
=
n
n
n
+ ( C n – 1 + C n) + C n]
( n Ci )( n C j )
=
n
= ( C0 + C1 +
2
Cr (31/2)256 – r (51/8)r
Cr 3128 – r/2 5r/8
This will be an integer if both r/2 and r/8 are integers,
and 0 < r < 256
r = 0, 8, 16, …, 256
i< j
n
256
256
n 2
+ Cn) = (2 ) = 2
2n
Thus, there are 33 integral terms.
Previous Years' AIEEE/JEE Main Questions
7. Note that S(1) is not true
x + y)n is 2n.
2n = 4096 = 212
12
2. ar =
p+q
ap =
Assume S(K) is true, then
n = 12
1 + 2 + … + (2K – 1) = 3 + K2
C6 = 924
1 + 2 + … + (2K – 1) + (2K + 1)
Cr
p+q
Cp =
p+q
= 3 + K2 + (2K + 1)
Cq = aq
= 3 + (K + 1)2
3. Let n = 0.0001 and
Ê 1ˆ
S = Á1 + ˜
Ë n¯
Thus, S(K)
n
Ê 1ˆ
Ê 1ˆ
= 1 + nC 1 Á ˜ + nC 2 Á ˜
Ë n¯
Ë n¯
=1+1+
+…+
Ê 1ˆ
+ … + nC n Á ˜
Ë n¯
n
8. Third term of (1 + a x)4 is the middle term and the 4th
term of (1 – a x)6
4
C2 a2 = 6C3 (– a)3
1 Ê 1ˆ 1 Ê 1ˆ Ê 2ˆ
Á1 - ˜ + Á1 - ˜ Á1 - ˜
2! Ë n ¯ 3! Ë n ¯ Ë n ¯
xn in (1 + x) (1 – x)n
= (1 + x) (–1)n (x – 1)n
= (–1)n (1 + x){nC0 xn – nC1 xn – 1 + …}
= (–1)n (nC0 – nC1) = (–1)n (1 – n)
1 Ê 1 ˆ Ê n - 1ˆ
˜
Á1 - ˜ Á1 n ¯
n! Ë n ¯ Ë
1 1
1
+ + +
n!
2! 3!
1
1
1
But
" r > 2.
=
£
r ! 1.2… r 2r -1
m
=1+
1 1
1
+ 2 + … + n -1
2 2
2
1 - (1/ 2)n
Ê 1ˆ
= 3-Á ˜
Ë 2¯
1 - 1/ 2
4. 2nCr + 1 =
2n
n -1
< 3.
C3r – 1
r + 1 = 3r – 1 or (r + 1) + (3r – 1) = 2n
r = 1 or n = 2r
As r > 1, n = 2r.
5. (r + 1)th term of (31/2 + 51/8)256 is
r + 1)th term in (1 + y)m is mCr. We
Cr – 1, mCr mCr + 1 are in A.P.
2(mCr) = mCr – 1 + mCr + 1
<1+1+
Thus, S < 1 + 1 +
6a2 = – 20a3
a = –3/10
1 n(n - 1) 1 n(n - 1)(n - 2)
+
2! n 2
3!
n3
1 n(n - 1) (n - n + 1)
n!
n2
=1+1+
+…+
2
S(K + 1)
m
2=
Cr -1
m
Cr
m
+
Cr +1
m
Cr
=
m!
r !(m - r )!
◊
(r - 1)!(m - r + 1)!
m!
+
m!
r !(m - r )!
◊
(r + 1)!(m - r - 1)!
m!
r
m-r
+
m - r +1 r +1
2(m – r + 1) (r + 1) = r(r + 1)
+ (m – r) (m – r + 1)
2=
2(mr – r2 + r + m – r + 1)
= r2 + r + m2 – 2mr + r2 + m – r
m2 – m(4r + 1) + 4r2 – 2 = 0
Mathematical Induction and Binomial Theorem
11. (r + 1)th term in the expansion of
Ê 2 1ˆ
ÁË ax + ˜¯
bx
11
Tr + 1 =
=
11
is
n – m = 10
Ê1ˆ
Cr (ax2)11 – r Á ˜
Ë bx ¯
Ê 1ˆ
ÁË ˜¯
b
Cr a11 – r
r
and n + m = 80
n = 45, m = 35
r
14. 20C0 –
x22 – 3r
+
x7, we set
7
x is
11
C5
20
a
20
b5
11
=
11
Ê 1 ˆ
Cr (ax)11 – r Á - 2 ˜
Ë bx ¯
Ê 1ˆ
Cr a11 – r Á - ˜
Ë b¯
20
C1 +
C2 – … +
20
C10 –
20
C11 + …
C20 = 0
2(20C0 –
6
1 ˘
È
Next, (r + 1)th term in the expansion of Íax = 2 ˙
bx ˚
Î
is
11
20
Using nCr = nCn – r, we get
22 – 3r = 7 or r = 5
tr + 1 =
1
[(n – m)2 – (n + m)]
2
and 10 = a2 =
11
7.45
20
C0 –
C1 +
20
20
C1 +
C2 – … +
20
20
C2 – … +
C10) –
20
C10 =
20
C10 = 0
1 20
( C10)
2
15. T5 + T6 = 0
n
r
C4 an – 4 (–b)4 + nC5 an – 5 (–b)5 = 0
a n - 4b4
a
r
n -5 5
b
=
n
C5
n
C4
a
n!
4!(n - 4)! n - 4
=
=
b 5!(n - 5)!
n!
5
x11 – 3r
x–7, we set 11 – 3r = – 7 or r = 6.
n
16. Â (r + 1) (nCr) xr
r =0
x–7 is
11
C5
11
a
C6
b5
b
6
=
=
11
n
n
= Â ( Cr )
r =0
6
b5
a6
a
5
C6
a
5
b6
a5
b6
[ ∵ nC r = nC n – r]
(ab)6 = (ab)5 or ab = 1
13. (1 – y)m (1 + y)n = 1 + a1y + a2y2 + …
(1 – y)m (1 + y)n
m(m - 1) 2
Ê
ˆ
y + ˜
= Á1 - my +
Ë
¯
2!
n(n - 1) 2
Ê
ˆ
y + ˜
¥ Á1 + ny +
Ë
¯
2!
1
= 1 + (n – m)y +
(m2 – m – 2mn + n2 – n)y2 +
2
…
Thus, 10 = a1 = n – m
d
{xr + 1}
dx
=
d Ï n n
r +1 ¸
Ì Â Cr x ˝
dx Ó r = 0
˛
=
d Ï Ê n n
r ˆ¸
Ì x Á Â Cr x ˜ ˝
¯˛
dx Ó Ë r = 0
d
{x(1 + x)n}
dx
= (1 + x)n + nx(1 + x)n – 1
=
Thus, Statement-2 is true.
Putting x = 1, we get
n
 (r + 1) (nCr) = 2n + n(1) 2n – 1 = (n + 2) 2n – 1
r =0
Thus, statemnen-2 is true and is a correct reason for
Statement-1.
17. 82n – 622n + 1
= (9 – 1)2n – (63 – 1)2n + 1
= (2nC0 92n – 2nC1 92n – 1 +
– 2nC2n – 1 (9) + 2nC2n)
– (2n + 1C0 (632n + 1) –
2n + 1
2n
C2 92n – 2 – …
C1 (632n)
Complete Mathematics—JEE Main
7.46
+
2n + 1
C2 (632n – 1) – … +
2n + 1
C1 (63) –
2n + 1
r7 – r
C 0)
For r = – 3, r7 – r = (–3)7 – (–3)
= 9m + 1 + 1 = 9m + 2 for some integer m
Thus, when 82n – 622n + 1
der is 2.
-
= (–3) [36 – 1]
= (–3) (128)
10
For r = –2, r7 – r = (– 2)7 – (– 2)
(1 + x)10 = Â (10 C j ) x j
j= 0
= – 128 + 2 = – 126
Differentiating both the side w.r.t. x, we get
10
10(1 + x)9 = Â j (10 C j ) x j -1
(1)
j =1
Again differentiating both the sides w.r.t, x, we get
Similarly, for r = – 1, 0, 1, 2, 3, r7 – r
by 7.
Thus, n7 – n
10
(10) (9) (1 + x)8 = Â j ( j – 1) (10Cj)xj – 2
(n + 7)7 – (n
(2)
j =1
(n + 1)7 – (n + 1) – (n7 – n
Putting x = 1 in (1) and (2), we get
(n + 1)7 – n7
10
S2 = Â j (10 C j ) = 10(29)
Thus, Statement-1 is true.
j =1
Statement-1 and Statement-2 are both true.
10
10
8
8
and S1 = Â j ( j - 1) ( Cj) = (10) (9) (2 ) = (90) (2 )
j =1
Hence Statement-2 is a correct explanation for Statement-1.
n
S3 = Â [ j + j ( j - 1)] (10Cj) = (10) (28) (2 + 9)
j =1
= 2 ÈÎ 2 n C1 ( 3 ) 2 n -1 + 2 n C3 ( 3 )2 n -3 + + 2 n C2 n -1 ( 3 )˘˚
S3 = (55) (29)
Thus, Statement-1is true but Statement-2 is false.
2
( 3 + 1) 2 n - ( 3 - 1) 2 n
3
2
19. 1 – x – x + x = (1 – x) – x (1 – x)
= 2 3 [2nC1 (3n – 1) +
2n
C3 (3n – 2) + … +
2n
C2n – 1]
This is an irrational number.
2
= (1 – x) (1 – x )
x – x2 + x3)6 =(1 – x)6 (1 – x2)6
7
2
3 6
x in (1 – x – x + x )
x
x7 in (1 – 6C1 x + 6C2 x2 – …
6
6
6
2
6
4
6
x +1
6
+ C6 x ) (1 – C1 x + C2 x – C3 x + …)
= (– 6C5) (– 6C1) + (– 6C3) (6C2)
2/3
and
= x1/3 + 1
- x1/ 3 + 1
x +1
1/ 2
x-x
=
x -1
x ( x - 1)
= 1 + x–1/2
+ (– 6C1) (– 6C3)
x +1
x -1 ˆ
Ê
+
(r + 1) term of ÁË 2 / 3
˜
1/ 3
x - x + 1 x - x1/ 2 ¯
= 36 – 300 + 120 = – 144
= (x1/3 – x–1/2)10 is
20. For Statement-2, write
10
th
Tr + 1 =
10
Cr (x1/3)10 – r (x–1/2)r
n = 7k + r where r = – 3, – 2, – 1, 0, 1, 2, 3.
Note that Tr + 1 will be independent of x if
Now n7 – n = (7k + r)7 – (7k + r)
10 - r r
- =0
3
2
7
7
n – n – (r – r) is a multiple of 7.
Thus, n7 – n
of x equals
10
r = 4. Thus, term independent
C4 = 210.
Mathematical Induction and Binomial Theorem
3 ˆ
23. T7 = T6 + 1 = 9C6 ÊÁ
Ë 841/ 3 ˜¯
a = 16
9-6
( 3 ln x)6
b = 272/3
28. (1 + x)101 (1 + x2 – x)100
Ê 33 ˆ
= (84) Á ˜ (33 (ln x)6) = 729
Ë 84 ¯
= (1 + x) (1 + x)100 (1 – x + x2)100
= (1 + x) (1 – x3)100
(ln x)6 = 1
ln x =
= (1 – x3)100 + x(1 – x3)100
x = e, 1/e
1
100
Thus, x = e.
= Â ( -1) r (100Cr) (x3r + x3r + 1)
r =0
24. See solution to Question 4.
25. Tr + 1 =
=
15
15
2 15 – r
Cr (x )
Ê 2ˆ
ÁË ˜¯
x
Thus, number of terms is 202.
r
E = (1 + x)1000 + x(1 + x)999 + x2(1 + x)998 + … +
Cr 2r x30 – 3r.
x1000
x15, set
È Ê x ˆ 1001 ˘
(1 + x)1000 Í1 - Á
˙
˜
˙˚
ÍÎ Ë 1 + x ¯
=
= (1 + x)1001 – x1001
1 - x / (1 + x)
r=5
30 – 3r = 15
and for term independent of x, set
r = 10
30 – 3r = 0
x50 in E
The required ratio is
15
C5 25 :
26. Tr + 1 =
=
10
15
10
5
C10 210 = 1 : 32
=
Cr (21/2)10 – r (31/5)r
Cr 2 2
–r/2
3
r/5
1001
C50 =
30. ÊÁ 2 + x ˆ˜
Ë
3¯
(0 < r < 10)
For rational terms, both r/2 and r/5
55
= Â
must be integers.
55
r =0
55
(1001)!
(50)!(951)!
55
=Â
r =0
55
Ê xˆ
Cr Á ˜
Ë 3¯
r
(255 – r)
Ê 255- r ˆ
Cr Á r ˜ x r
Ë 3 ¯
This is possible if r = 0 or r = 10
Thus, required sum
=
10
5
C0 2 +
10
are equal
5
C10 2 2
–5
2
3 = 41.
55
27. (1 + ax + bx2) (1 – 2x)18
= (1 + ax + bx2) [1 –
–
18
C3 (2x)3 +
18
18
C1 (2x) +
18
=–
3
C4 (2x)4 – …]
18
2
18
18
C4 24 –
b=–
18
18
C2 (22) (b) = 0
544
+ 17a
3
and b = – 80 +
32
a
3
- 544
32
+ 17a = – 80 +
a
3
3
Ê 254 - r ˆ
Cr + 1 Á r +1 ˜
Ë 3 ¯
55!
r !(55 - r )!
◊
(r + 1)!(54 - r )!
55!
6=
55 - r
r +1
6r + 6 = 55 – 4
C3 (23) (a) +
55
6=
C3 (2 ) + ( C2)2 a – ( C1) (2) b = 0
x4
=
Ê 255- r ˆ
Cr Á r ˜ =
Ë 3 ¯
C2 (2x)2
x3
18
7.47
r=7
Thus, these terms are 8th and 9th.
31. Replacing x by x – 1, we get
5
i
 ai (1 + x - 1) = 1 + (x – 1)4 + (x – 1)5
i =0
5
fi  ai xi = 1 + (1 – x)4 – (1 – x)5
i= 0
Complete Mathematics—JEE Main
7.48
x2 on RHS
a2
Tr + 1 = Cr(2x )
= 4C2 – 5C2 = 6 – 10 = – 4.
Â
32. (1 + xn + x253)10 =
r ,s≥0
r + s £ 10
2 8–r
8
Ê 1ˆ
ÁË - ˜¯
x
= 8Cr (–1)r 28 – r x16 – 3r
10! x 253r x ns
r ! s !(10 - r - s )!
To obtain term independent of x in the expansion of
1ˆ
Ê 1
5ˆ Ê
2
ÁË1 - + 3x ˜¯ ÁË 2 x - ˜¯
x
x
253r + ns = 1012
10!
4 !0 !6 !
+
50
50
50
(
=
=
=1+
50
3
50
C3 ( -2 x ) +
r=7
and 16 – 3r = – 5
50
C1 ( -2 x ) +
C4 ( -2 x )
50
C2 ( -2 x )
2
50
50
= (56) (8) – 3(8) (2) = 400
n
)
C2 (– 2)2 +
= (– 1) (8C5) (– 1)5 28 – 5 + 3(8C7) (– 1)7 28 – 7
4
C49 (-2 x )49
C50 -2 x
=1+
r=5
Thus, term independent of x in (1)
+…+
+
16 – 3r = 1
C4
33. (1 - 2 x )
50
(1)
But 16 – 3r = 0 has no solution;
\
=
8
we set 16 – 3r = 0, 16 – 3r = 1 and 16 – 3r = – 5
fi r = 4, s = 0.
10
r
C4 (– 2)4 + … +
50
1
[(1 + (– 2))50 + (1 – (– 2))50]
2
C50 (– 2)50
Ê 2 4ˆ
36. Number of terms in the expansion of ÁË1 - + 2 ˜¯
x x
1
is n + 2C2 =
(n + 2) (n + 1)
2
1
(n + 1) (n + 1) = 28
n=7
2
x=1
+ 4)7 = 37 = 2187.
1
(350 + 1)
2
2016
37. Â a j xi = (1 + x)2016 + x(1 + x)2015 + x2(1 + x)2014
r + 1)th term.
i= 0
+ … + x2016
È Ê x ˆ 2017 ˘
(1 + x) 2016 Í1 - Á
˙
˜
˙˚
ÍÎ Ë 1 + x ¯
=
x
11+ x
tr + 1 = nC r x r
n
Cr : nCr + 1 : nCr + 2 = 1 : 7 : 42
n
n
Cr
Cr +1
=
1
and
7
n
n
Cr +1
Cr + 2
=
7
42
= (1 + x)2017 – x2017
r +1 1
1
r+2
=
and
=
n-r 7
n - r -1 6
n = 8r + 7 and n = 7r + 13
8r + 7 = 7r + 13
=
C17 =
term.
38. Tr + 1 =
35. General term, Tr + 1
=
8
2017
r=6
th
Ê 2 1ˆ
of ÁË 2 x - ˜¯
x
x17
a17
18
18
2017 !
17 !2000!
Ê 1 ˆ
Cr (x1/3)18 – r Á 1/ 3 ˜
Ë 2x ¯
Ê 1ˆ
Cr x6 – 2r/3 Á ˜
Ë 2¯
is
r
r
x–2, 6 – 2r/3 = – 2
r = 12
8 = 2r/3
Mathematical Induction and Binomial Theorem
n
–4
x ,
6 – 2r/3 = – 4
= Cn – 6 (2 )
10 = 2r/3
12
m=
Thus,
18
Ê 1ˆ
C12 Á ˜ , n =
Ë 2¯
m
=
n
18
18
C12
18
r = 15
Ê 1ˆ
C15 Á ˜
Ë 2¯
=
18! 15!3! 3
(2 )
◊
12!6! 18!
=
15 ¥ 14 ¥ 13
◊ (8) = 182
6¥5¥4
n
1/3 6
= C6 (2 ) (3
)
–1/3 n – 6
a 1
=
b 6
(21/ 3 )n - 6 (3-1/ 3 )6
1/ 3 6
-1/ 3 n - 6
(2 ) (3
)
Ê (21/ 3 ) ˆ
Á -1/ 3 ˜
Ë (3 ) ¯
)
=
1
6
=
1
6
n -12
6(n – 12)/3 = 6–1
n – 12 = – 3
Previous Years’ B-Architecture Entrance
Examination Questions
Ê 1ˆ
1. Tr + 1 = nCr (x3)n – r Á - 2 ˜
Ë x ¯
(3
r
= nCr (–1)r x3n – 5r
n = 9.
4. 7128 = (72)64 = (50 – 1)64 = (1 – 50)64
64
=1+ Â
64
Ck ( -50)k
k =1
Thus, remainder when 7128
5. For 0 < j < 8,
x5, set 3n – 5r = 5
r = (3n – 5)/5 = l (say)
1
( 8C j)
( j + 1)( j + 2)
=
8!
( j + 1)( j + 2) j !(8 - j )!
Note that l – m = 1.
=
1
10!
10 ¥ 9 ( j + 2)!(10 - ( j + 2))!
n
=
1 10
( C j +2 )
90
x10, set 3n – 5r = 10
r = (3n – 10)/5 = m (say)
Cl (–1)l + nCm (–1)m = 0
n
C l = nC m
l + m = n.
3n - 5 3n - 10
=n
+
5
5
6n – 15 = 5n
n = 15.
=
15
Thus,
S=
x in
(1 + x) (1 + x + x2) (1 + x + x2 + x3)
… (1 + x + x2 + … + xn)
x in (1 + x)n
ˆ
1 Ê 10 10
 C j +2 ˜
Á
90 Ë j = 0
¯
=
1 Ê 10 10
ˆ
 Ck -10C0 -10C1 ˜
Á
¯
90 Ë k = 0
=
1
1013
(210 – 1 – 10) =
90
90
C2 (–1)2 = 105
2. a1
7.49
–1/3 n – 6
15
(23)
C15
1/3 n – (n – 6)
n
6. Â a n - j b j = an + an – 1 b + an – 2 b2 + … +
j =0
abn – 1 + bn
n
= C1 = n
th
3. Let a = 7 term from the beginning
=
a n [1 - (b /a)n +1 ]
[sum of a G.P.]
1 - b /a
=
a n +1 - b n +1
a-b
= T7 = T6 + 1 = nC6 (21/3)n – 6 (3–1/3)6
th
and let b = 7 term from the end
= (n + 2 – 7)th or (n – 5)th term from the beginning
xm in (a x + b)m is (m + 1) am b
Complete Mathematics—JEE Main
7.50
x + y)n
Statement-2 is true.
is 2n. Therefore,
n
Next, Â (-1)
j
(2x + 3)
n–j
(5 – 2x)
j
2n = 2048 = 211
j =0
n = 11.
n
As n
n
= Â (2 x + 3)
n- j
(2x – 5)
j
Cr is nCk where
1
1
(n – 1) or k =
(n + 1).
2
2
Here n = 11, therefore k = 5 or 6.
k=
j =0
=
(2 x + 3)n +1 - (2 x - 5)n +1
(2 x + 3) - (2 x - 5)
=
1
[(2x + 3)n + 1 – (2x – 5)n + 1]
8
11
10. (1 + x + x2)8 =
p,q,r ≥0
P + q + r =8
xn is
1 n+1
[
Cn 2n (3) –
8
Â
C 6.
8!
1p xp(x2)r
p !q !r !
For a5, we require, q + 2r = 5,
n+1
Cn 2n (–5)]
q = 5, r = 0, p = 3 or q = 3, r = 1, p = 4,
or q = 1, r = 2, p = 5.
= (n + 1) 2n
Statement-1 is true and Statement-2 is correct explanation for it.
7. xn = [(1 + x) – 1]n = [1 – (1 – x)]n
x5 is
8!
8!
8!
= 504
+
+
5!3! 3!1!4! 1!2!5!
t24 in (1 + t2)12 (1 + t12) (1 + t24)
n
= Â n Ck (1 + x)n – k (–1)k
t24 in (1 +
k =0
12
C1 t2 +
= Â n Ck (–1)k (1 – x)k
=
k =0
12
C12 +
12
C6 + 1 =
12
C6 + 2
12. Let
x)k in
ak
x)59
S
n
n–k
 Ck (1 + x)
k
n–k n
(–1) = (–1)
k =0
101
C51)
C30 +
59
C31 + … +
59
C59
(1)
n
S=
59
C29 +
59
C28 + … +
59
C0
(2)
Adding (1) and (2), we get
59
C0 +
59
C1 + … +
59
C59 = 259
= (1 – mx + mC2 x2 – …) (1 + nx + nC2 x2 + …)
= 100
(2a – 3)a = 2 where a = log10 x
2a2 – 3a – 2 = 0
= 1 + (n + m)x + (mC2 – mn + nC2)x2 + …
n – m = 3, mC2 – mn + nC2 = – 4
(2a + 1) (a – 2) = 0
As x > 1, a = log10 x > 0, thus,
log10 x = 2
59
(1 – x)m (1 + x)n
As T3 = 1000, we get
a=2
=
S = 258.
= 10 x 2 log10 x -3
2 log10 x - 3
Ck
2S =
3
8. T3 = C2 ÊÁ 1 ˆ˜ ( x log10 x )2
Ë x¯
5
Cn – k = (–1)
Using Cr = Cn – r, we get
For n = 101, k = 51, we get
(a51, b51) = (101C51, –
n–k n
n
and bk = nCk (– 1)k
x
C2 t4 + …)
(1 + t12 + t24)
n
n
12
x = 100
1
(m2 – 2mn + n2 – (m + n)) = – 4
2
n – m = 3, (m – n)2 – (m + n) = – 8
n – m = 3,
Mathematical Induction and Binomial Theorem
n – m = 3, m + n = 17
n = 10, m = 7
m : n = 7 : 10
(a + b)2n – 3 is
7.51
1
[2n – 3C0 + 2n – 3C1 + … + 2n – 3C2n – 3]
2n - 3 + 1
1
22( n - 2)
=
22n – 3 =
= 16 = 24
2(n - 1)
n -1
n=5
CHAPTER EIGHT
Progressions
SEQUENCES
A sequence is a function whose domain is the set of natural
numbers. If the range of a sequence is a subset of real numbers (complex numbers), then it is called a real sequence
(complex sequence).
If f : N Æ C, is a sequence, we usually denote it by
· f (1), f (2), f (3), Ò = · f (n) Ò
If tn = f (n), then the sequence is written as · t1, t2, t3, Ò
ARITHMETIC PROGRESSION
A sequence of numbers ·tnÒ is said to be in arithmetic
progression (A.P.) when the difference tn – tn–1 is a constant
for all n ΠN, n > 1. This constant is called the common difference of the A.P., and it is usually denoted by the letter d.
If a is the first term and d the common difference of the
A.P., then its nth term tn is given by
tn = a + (n – 1)d
tn = tm + (n – m)d
The sum Sn of the first n terms of such an A.P. is given by
n
n
Sn = [2a + (n – 1)d] = (a + l)
2
2
where l is the last term, that is, the nth term of the A.P.
Some Facts About A.P.
1. If a fixed number is added to (subtracted from) each
term of a given A.P., then the resulting sequence is
also an A.P., and it has the same common difference
as that of the given A.P.
2. If each term of an A.P. is multiplied by a fixed
constant (or divided by a non-zero fixed constant),
then the resulting sequence is also an A.P.
and b1, b2, b3,
are two arithmetic
3. If a1, a2, a3,
progressions, then a1 + b1, a2 + b2, a3 + b3, is
also an A.P.
4. If a1, a2, a3, º are in A.P., then
am – an = (m – n)d
where d is the common difference of the A.P.
Illustration
1
If 20th term of an A.P. is 25 and 25th term is 15, find 17th
term of the A.P.
If d is the common difference of the A.P., then
(25 – 20)d = a25 – a20 = 15 – 25
fi d = – 2.
Also a17 – a20 = (17 – 20)d = 6
fi a17 = 25 + 6 = 31
5. If we have to take three terms in A.P. whose sum
is known, it is convenient to take them as a – d,
a, a + d. In general, if we have to take (2r + 1)
terms in A.P., we take them as a – rd, a – (r – 1)d,
, a – d, a, a + d, , a + rd.
6. If we have to take four terms in A.P., whose sum is
known, we take them as a – 3d, a – d, a + d, a + 3d.
7. The arithmetic mean A of any two numbers a and
b is given by (a + b)/2. It is to be noted that the
sequence a, A, b is in A.P.
If a1, a2, , an are n numbers, then the arithmetic
mean A, of these numbers is given by
A=
1
(a1 + a2 +
n
+ a n)
8. The n numbers A1, A2, , An are said to be n arithmetic means between a and b if a, A1, A2, , An, b is
an A.P. Here a is the first term and b is the (n + 2)th
term of the A.P. If d is the common difference of
this A.P., then
b-a
b = a + (n + 2 – 1)d = a + (n + 1)d fi d =
n +1
Thus,
A1 = a +
b-a
2(b - a )
, A2 = a +
,
n +1
n +1
An = a +
n(b - a )
n +1
8.2
Complete Mathematics—JEE Main
TIP
Sometimes, it is useful to use An = b – d etc.
Sum of n Arithmetic means A1 + A2 + . . . + An =
n
(a + b)
2
GEOMETRIC PROGRESSION
A sequence is said to be in geometric progression (G.P.)
when its first term is non-zero and each of its succeeding
terms is r times the preceding term, where r is some fixed
non-zero number. The fixed number r is known as the common ratio of the G.P. It is clear from the definition that no
term of a G.P. is zero. If a is the first term of a G.P. and r its
common ratio, then its nth term, tn, is given by
tn = ar n – 1
The sum Sn of the first n terms of such a G.P. is given by
n
Ï a(r - 1)
Ô
Sn = Ì r - 1
Ô na
Ó
If –1 < x < 1, then
if r π 1
.
if r = 1
lim x n = 0.
nƕ
Therefore, the sum of the infinite G.P.
1
=
.
1 + x + x2 +
1- x
(–1 < x < 1)
(1)
Thus (1 – x)–1= 1 + x + x2 + x3 +
Differentiating both the sides w.r.t. x, we get
(1 – x)–2 = 1 + 2x + 3x2 + 4x3 + (–1 < x < 1) (2)
Differentiating (1) (m – 1) times w.r.t. x, we get for
– 1< x < 1
(m + 1)! 2
(m – 1)! (1 – x)–m = (m – 1)! + m!x +
x
2!
(m + 2)! 3
+
x +
3!
fi
(1 – x)–m = 1 + mC1 x + m +1C2 x2 + m +2C3 x3 +
(3)
for – 1 < x < 1
Also, if n is any real number, then
(1 – x)–n = 1 + nx +
n (n + 1) 2
n (n + 1)(n + 2) 3
x +
x
3!
2!
+
(–1 < x < 1)
If – 1 < r < 1, then the sum of the infinite G.P.
a
a + ar + ar2 +
=
1- r
Some Facts About G.P.
1. If each term of a G.P. is multiplied (divided) by a
fixed non-zero constant, then the resulting sequence
is also a G.P.
2. If a1, a2, a3,
and b1, b2, b3,
are two geometric
progressions, then the sequence a1b1, a2b2, a3b3,
is also a G.P.
3. The geometric mean G of two positive numbers a
and b is given by a b . It is to be noted that a,
G and b are in G.P. If a1, a2, , an are n positive
numbers, then their geometric mean is given by
G = (a1 a2 an)1/n
4. The n numbers G1, G2, . . . Gn are said to be n
geometric means between two positive real numbers
a and b if a, G1, G2, . . ., Gn, b forms a G.P. Let
r be the common ratio of this G.P., then
b = (n + 2)th term of the G.P.
= ar n + 2 – 1 = ar n + 1
1 ( n + 1)
Ê bˆ
fir= Ë ¯
a
1 ( n + 1)
2 ( n + 1)
Ê bˆ
Ê bˆ
Thus, G1 = a Ë ¯
, G2 = a Ë ¯
a
a
n ( n + 1)
Ê bˆ
. . ., Gn = a Ë ¯
a
Also, note that
G1 G2 . . . Gn = (ab)n / 2
5. If we have to take three terms in G.P., we take them
as a/r, a, ar and four terms as a/r3, a/r, ar, ar3.
is a G.P. (ai > 0 " i), then log a1,
6. If a1, a2, a3,
is an A.P. In this case the conlog a2, log a3,
verse also holds.
is a G.P. Let ai = AR i – 1,
Suppose a1, a2, a3,
where A is the first term and R is the common ratio
of the G.P. Then
log ai = log A + (i – 1) log R
is
This shows that log a1, log a2, log a3,
an A.P. whose first term is log A and the common difference is log R. Conversely, assume that
is an A.P. Suppose that
log a1, log a2, log a3,
log ai = a + (i – 1)d, where a is the first term and
d is the common difference of the A.P. Then
ai = a a + (i – 1)d = a a (a d)i – 1
where a is the base of the logarithm. This shows
is a G.P., with the first term a a
that a1, a2, a3,
and common ratio a d.
ARITHMETICO-GEOMETRIC
SEQUENCE
Suppose a1, a2, , an,
is an A.P., and b1, b2, , bn,
is a G.P. Then the sequence a1 b1, a2 b2, , an bn, is said
to be an arithmetico-geometric sequence. An arithmeticogeometric sequence is of the form
ab, (a + d)br, (a + 2d)br 2, (a + 3d)br 3,
Sum to n term of an Arithemtico-Geometric series.
Let Sn = ab + (a + d)br + (a + 2d)br2 + +
[a + (n – 2)d]br n–2 + [a + (n – 1)d]br n –1
(1)
Progressions
Thus
We multiply each term by r and write the first term below
the second, the second below the third and so on.
rSn = abr + (a + d)br2 + + [a + (n – 3)d]br n – 2
+ [a + (n – 2)d]br n – 1 + [a + (n – 1)d]br n
Subtracting, we get
(1 – r)Sn = ab + dbr + dbr2 +
+ dbr n – 2 + dbrn – 1
– [a + (n – 1)d]br n
= ab +
fi Sn =
dbr (1 - r n - 1 )
– [a + (n – 1)d]br n
1- r
a b dbr (1 - r n - 1 ) [ a + (n - 1)d ]br n
+
,rπ1
1- r
1- r
(1 - r )2
1
1
a-b
= +
,
H1 a (n + 1)ab
1
1
1 n( a - b )
1 2(a - b)
= +
= +
, ,
H2
H n a (n + 1) ab
a (n + 1) ab
From here, we can get the values of H1, H2,
H n.
lim r = 0
nƕ
n
1.
Âk
1
n (n + 1)
2
=1+2+3+º+n=
k =1
and
n
n
lim nr = 0
2.
nƕ
Therefore, the sum of an infinite number of terms of the
series in (1) is
ab
dbr
+
(–1 < r < 1)
S=
1 - r (1 - r )2
Â
k2 = 12 + 22 + º + n2 =
k =1
n
3.
1
1
1
1
where a =
and d =
an =
a + (n - 1) d
a1
a2 a1
Â
k =1
3
k =1
n
4. To find
1 Ê 1 1ˆ
a-b
fid=
Á - ˜=
n + 1 Ë b a ¯ (n + 1) ab
n
Â
k4 and
Â
k =1
5
k5 we can use the identities:
k =1
k – (k – 1) = 5k –10k3 + 10k2 – 5k + 1
and k6 – (k – 1)6 = 6k5 – 15k4 + 20k3 – 15k2 +
6k – 1
The first one gives
5
4
n
Â
n
{k5 – (k – 1)5} = 5
k =1
n
k4 – 10
Â
k =1
Â
k3 +
k =1
n
10
Â
n
k2 – 5
k =1
fi
n5 + 10
1 1Ê 1
1
1ˆ
= Á +
++ ˜
H n Ë a1 a2
an ¯
1 1
= + (n + 2 – 1)d
b a
2
3
2
1 2
È1
˘
= Í n ( n + 1)˙ =
n (n + 1)2
4
Î2
˚
1. If a and b are two non-zero numbers, then the
harmonic mean of a and b is a number H such that
the sequence a, H, b is a H.P. We have
2. The n numbers H1, H2, , Hn are said to be harmonic means between a and b if a, H1, H2, , Hn,
b are in H.P., that is, if 1/a, 1/H1, 1/H2, , 1/Hn,
1/b are in A.P. Let d be the common difference of
this A.P. Then
3
= (1 + 2 + 3 + º + n)2
Some Facts about H.P.
2 ab
1 1 Ê 1 1ˆ
or
H=
= Á + ˜
Ë
¯
a
+b
H 2 a b
If a1, a2, , an are n non-zero numbers, then the
harmonic mean H of these numbers is given by
1
n(n + 1) (2n + 1)
6
Ê n ˆ
k = 1 + 2 + º + n = Á Â k˜
Ë
¯
3
HARMONIC PROGRESSION
The sequence a1, a2, , an,
where ai π 0 for each i is
said to be in harmonic progression (H.P.) if the sequence
1/a1, 1/a2, , 1/an,
is in A.P. Note that an, the nth term
of the H.P., is given by
,
SUMMATION OF SOME SERIES OF
NATURAL NUMBERS
If –1 < r < 1, then
n
8.3
{
1 2
2
n ( n + 1)
4
} {
– 10
Â
k+n
k =1
}
1
n ( n + 1) (2n + 1)
6
1
+5 n (n + 1) - n
2
{
n
}
= 5Â k 4
k =1
n
Similarly we can use 2nd identity to obtain sum of
 k5 .
k =1
SUM OF THE PRODUCTS OF TWO
TERMS OF A SEQUENCE
To obtain the sum
Â
ai aj, we use the identity
i< j
2
Â
i< j
ai aj = (a1 + a2 + º + an)2 – (a 21 + a 22 + º + a 2n )
Complete Mathematics—JEE Main
8.4
where
f (r) = (2r – 3)tr
= (2r – 3) (2r – 1) (2r + 1) (2r + 3)
= (4r2 – 9) (4r2 – 1)
This gives
METHOD OF DIFFERENCE FOR
SUMMATION OF SERIES
If possible express rth term as difference of two terms as
follows:
tr = f(r) – f(r ± 1).
Illustration
n
8
n
Â
tr =
r =1
Â
{f(r + 1) – f (r)} = f(n + 1) – f (1)
r =1
n
2
fi
8
Â
tr = [4(n + 1)2 – 9] [4(n + 1)2 – 1]
r =1
If the given series is 1.3.5 + 3.5.7 + 5.7.9 + º
then tr = (2r – 1) (2r + 1) (2r + 3)
and tr +1 = (2r + 1) (2r + 3) (2r + 5)
fi (2r + 5)tr = (2r – 1)tr +1
fi (2r – 3)tr + 8tr = (2r – 1)tr +1
fi 8tr = f (r + 1) – f (r)
– (4 – 9) (4 – 1)
= 16(n + 1)4 – 40(n + 1)2 + 9 + 15
n
fi
Â
tr = 2(n + 1)4 – 5(n + 1)2 + 3
r =1
SOLVED EXAMPLES
Concept-based
Straight Objective Type Questions
Example 1: Suppose a1, a2, . . . are in A.P. If a8 : a5 =
3 : 2, then a17 : a23 is:
(a) 1 : 2
(b) 3 : 4
(c) 4 : 11
(d) 8 : 11
Ans. (b)
Solution: Let a be the first term and d be the common
difference of the A.P. a1, a2, a3, . . . Then
a
3
a + 7d
= 8 =
a5 a + 4d
2
fi
3a + 12d = 2a + 14d fi a = 2d.
Now,
a17
a + 16d 2d + 16d 18 3
=
=
=
=
a23 a + 22d 2d + 22d 24 4
Example 2: If 7th term of an A.P. is 9 and 9th term of
the A.P. is 7, then 20th term of the A.P. is
(a) – 2
(b) – 3
(c) – 4
(d) – 6
Ans. (c)
Solution: If d is common difference of the A.P., then
2d = a9 – a7 = 7 – 9 = – 2
fi
d = – 1.
\
a20 = a9 + (20 – 9)d = 7 – 11 = – 4
Example 3: Suppose mth term of an A.P. is 1/n and nth
term of the A.P. is 1/m. If rth term of the A.P. is 1, then r is
equal to
(a) mn
(b) m + n
(c) m – n
(d) m + n – 1
Ans. (a)
Solution: Let d be the common difference of the A.P.,
then
1 1
(m – n)d = am – an = n m
1
fi
d=
mn
Now
ar – am = (r – m)d
fi
1 = ar =
fi
r = mn.
r
1
1
=
+ (r – m)
n
mn mn
Example 4: The number of terms of the A.P. 1, 4, 7, . . .
that must be taken to obtain a sum of 715 is
(a) 24
(b) 23
(c) 22
(d) 21
Ans. (c)
Solution: Here a = 1, d = 3 and Sn = 715. We have
n
n
715 = [2a + (n – 1)d] = [2(1) + (n – 1)(3)]
2
2
2
2
fi
1430 = 3n – n or 3n – n – 1430 = 0
1
(1 ± 1 + 3 ¥ 4 ¥ 1430 ) = 1 (1 ± 131)
fi
n=
6
6
= 22, –65/3
As n must be a natural number n = 22
Example 5: If sum of first 20 terms of an A.P. is equal
to sum of first 30 terms of the A.P. then sum of the first 50
terms of the A.P. is
Progressions
(a) – 1
(c) 10
Ans. (b)
(b) 0
(d) 25
Solution: Let a be the first term and d be the common
difference of the A.P., then
S20 = S30
20
30
[2a + (20 – 1)d] =
[2a + (30 – 1)d]
2
2
fi
2a + 49d = 0
50
Now,
S50 =
[2a + (50 – 1)d] = 0.
2
Example 6: If for every n ΠN, sum to n terms of an A.P.
is 5n2 + 7n then its 10th term is
(a) 7/2
(b) 570
(c) 102
(d) 52
Ans. (c)
fi
Solution: We have
ar = Sr – Sr – 1
fi
a10 = S10 – S9
= 5 (102 –92) + 7 (10 – 9)
= 102
Example 7: If the sum of three numbers in A.P. is 24
and their product is 440, then common difference of the
A.P. can be
(a) 3
(b) 2
(c) 5
(d) – 5
Ans. (a)
Solution: Let the three number in A.P. be a – d, a, a + d.
Then
24 = (a – d) + a + (a + d) = 3a
fi
a=8
Also,
440 = (a – d) a(a + d) = 8(64 – d2)
fi
d2 = 64 – 55 = 9 fi d = ± 3.
\
d can be 3.
Example 8: The digits of a three digit number N are in
A.P. If sum of the digits is 15 and the number obtaind by reversing the digits of the number is 594 less than the original
1000
is equal to
number, then
N - 252
(a) 5/6
(b) 5/3
(c) 0.06
(d) 0.03
Ans. (b)
Solution: Let three digits of N be:
Hundred’s digit = a – d
ten’s digit = a,
and unit’s digit = a + d
(a – d) + a + (a + d) = 15 fi a = 5
and
N = 100(5 – d) + 50 + (5 + d)
Number obtained by reversing digits of N is
N1 = 100(5 + d) + 50 + (5 – d)
Now,
594 = N – N1 = 100(–2d) + 2d
fi
2d = – 6 fi d = – 3
Thus,
N = 852 and
1000 5
1000
=
=
600 3
N - 252
8.5
(1)
(2)
Example 9: If sum of four numbers in A.P. is 28 and
product of two middle terms is 45, then product of the first
and last terms is
(a) 11
(b) 13
(c) 15
(d) 17
Ans. (b)
Solution: Let four numbers in A.P. be
a – 3d, a – d, a + d, a + 3d
We are given
28 = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 4a
fi
a=7
Also,
(a – d) (a + d) = 45
fi
49 – d2 = 45 fi d = ± 2.
Now,
(a – 3d) (a + 3d) = a2 – 9d2 = 13
Example 10: n arithmetic means are inserted between
3 and 17. If the ratio of first to last mean is 1 : 3, then n is
equal to:
(a) 4
(b) 5
(c) 6
(d) 8
Ans. (c)
Solution: Let the n arithmetic means between 3 and
17 be A1, A2, . . . , An. If d is the common difference, then,
d=
Also,
fi
17 - 3
14
=
n +1 n +1
A1
1
3+d 1
=
fi
=
An
3
17 - d 3
3 + 14 ( n + 1)
1
3n + 17 1
= fin=6
= fi
17 - 14 (n + 1)
3
17n + 3 3
Example 11: Suppose a, b, c > 0 and p ΠR. If
(a2 + b2)p2 – 2(ab + bc)p + (b2 + c2) = 0
(1)
then a, b, c are in
(a) A.P
(b) G.P
(c) H.P.
(d) A.G.P.
Ans. (b)
Solution: Write (1) as
(a2 p2 – 2abp + b2) + (b2p2 – 2bcp + c2) = 0
fi
(ap – b)2 + (bp – c)2 = 0
As
ap Рb, bp Рc ΠR.
8.6
Complete Mathematics—JEE Main
from (2), we get
ap – b = 0, bp – c = 0
b c
fi
p= =
a b
fi
a, b, c are in G.P.
Example 14: Sum to 25 terms of the series 0.5 + 0.55
+ 0.555 + . . . is:
Example 12: Suppose (m + n)th term of a G.P. is p and
(m – n)th term is q, then its nth is
m n
Ê qˆ
pq
(b) p Á ˜
(a)
Ë p¯
m 2n
Ê qˆ
(c) p Á ˜
Ë p¯
5
(224 – 10–25)
81
5
(224 – 10–24)
(c)
81
(a)
(d)
p
q
Ans. (c)
Now,
am + n
p
Ê pˆ
=
= r2n fi r = Á ˜
Ë q¯
am - n
q
an = arn – 1 =
Ê pˆ
= pÁ ˜
Ë q¯
ar
-m 2n
m + n -1
Ê qˆ
= pÁ ˜
Ë p¯
3069
3 (1 - (1 2 )n )
= Sn =
512
1-1 2
n
fi
1
Ê1 ˆ
Ë 2 ¯ = 1024 fi n = 10
5
[(1 – 0.1) + (1 – (0.1)2) + (1 – (0.1)3)
9
+ . . . + (1 – (0.1)25)]
=
(0.1) (1 - (0.1)25 ) ˘
5 È
Í25 ˙
9 Î
1 - 0.1
˚
=
5
[224 – 10–25]
81
Solution: Let three numbers in A.P. be a – d, a, a + d.
As
21 = (a – d) + a + (a + d) = 3a
fi
a=7
Also,
7 – d, 7 – 1, 7 + d + 1 are in G.P.
fi
62 = (7 – d) (8 + d)
fi
36 = 56 – d – d2
fi
d2 + d – 20 = 0
fi
(d + 5) (d – 4) = 0 fi d = – 5, 4
As,
7 – d, 6, 8 + d is an increasing G.P., d = 4. The
6
6
common ratio of the G.P. is
=
= 2.
7-4
7-d
Solution: Here a = 3, r = 1/2.
Let the number of terms needed be n, then
1023
Ê1 ˆ
= 1- Ë ¯
1024
2
=
Example 15: The sum of three numbers in A.P. is 21. If
the second number is reduced by 1 while third is increased
by 1, three resulting numbers form an increasing G.P., the
common ratio of the G.P. is
(a) 2
(b) 3
(c) 4
(d) 7
Ans. (a)
m 2n
Example 13: The number of terms of the G.P.
3 3
3069
is:
3, , , . . . needed to obtain a sum of
2 4
512
(a) 9
(b) 10
(c) 11
(d) 12
Ans. (b)
fi
(d) none of these
Solution: Let
S = 0.5 + 0.55 + 0.555 + . . . + upto 25 terms
5
= [0.9 + 0.99 + 0.999 + . . . + upto
9
25 terms]
1 2n
rm
5
(224 – 10–25)
9
Ans. (a)
Solution: Let a be the first term and r be the common
ratio. We have
p = am + n = arm + n – 1
q = am – n = arm – n – 1
fi
(b)
n
LEVEL 1
Straight Objective Type Questions
Example 16: If one geometric G and two arithmetic
means A1 and A2 are inserted between two distinct positive
numbers, then
Ê 2 A1 - A2 ˆ Ê 2 A2 - A1 ˆ
Ë
¯Ë
¯
G
G
Progressions
is equal to
(a) 0
(c) – 1.5
Ans. (b)
(b) 1
(d) – 2.5
Solution: Let two distinct positive numbers be a and b.
Then G2 = ab
Let common difference of the A.P. a, A1, A2, b. be d.
We have
2A1 – A2 = 2(a + d) – (a + 2d) = a
and 2A2 – A1 = 2(b – d) – (b – 2d) = b
ab
Ê 2 A - A2 ˆ Ê 2 A2 - A1 ˆ
= 2 =1
Thus, Ë 1
¯
Ë
¯
G
G
G
Example 17: If sum of the infinite G.P.
1
1
p + 1 + + 2 + . . . , (p > 2) is 49/6, then sum of the 3rd
p p
term and the 4th term of the G.P. is
8
1
(a)
(b)
49
7
6
23
(c)
(d)
7
49
Ans. (a)
Solution: For p > 1,
49
1
1
p
=p+1+ + 2 +...=
p p
6
1 - 1/ p
2
fi
fi
fi
fi
As
49
p
=
6
p -1
6p2 – 49p + 49 = 0
6p2 – 42p – 7p + 49 = 0
(6p – 7) (p – 7) = 0
p > 2, p = 7
2
3
Ê 1ˆ
Ê 1ˆ
Thus, a3 + a4 = p Á ˜ + p Á ˜
Ë p¯
Ë p¯
1 1
8
=
= +
7 49 49
Example 18: If a1, a2, a3, . . . , an, . . . are in A.P. such
that a4 – a7 + a10 = m, then sum of the first 13 terms of the
A.P. is
(a) 10 m
(b) 12 m
(c) 13 m
(d) 15 m
Ans. (c)
Solution: Let d be the common difference of the A.P.,
then
m = a4 – a7 + a10 = a4 + (a10 – a7)
= a1 + 3d + 3d = a1 + 6d
Now,
13
[2a1 + (13 – 1)d] = 13m.
S13 =
2
8.7
Example 19: If (20)19 + 2(21) (20)18 + 3(21)2 (20)17
+ . . . + 20(21)19 = k (20)19
then k is equal to
(a) 400
(b) 100
(c) 441
(d) 420
Ans. (a)
Solution: Dividing by (20)19, we can write
k = 1 + 2x + 3x2 + . . . + 20x19
where x = 21/20.
d
k=
[x + x2 + . . . + x20]
dx
=
=
d È x ( x20 - 1) ˘
Í
˙
dx Î x - 1 ˚
( x - 1) (21x 20 - 1) - x( x 20 - 1)
( x - 1)2
21
20
= 20 x - 21x + 1
( x - 1)2
20x21 – 21x20
But
21
21
Ê 21 ˆ
Ê 21 ˆ
= 20 Ë ¯ - 21Ë ¯ = 0
20
20
1
k=
= 400
(21 20 - 1)2
\
Example 20: Three positive numbers form an increasing G.P. If the middle term in this G.P. is tripled, the new
numbers are in A.P. Then the common ratio of G.P.is:
(a) 3 + 2 2
(b) 2 2 - 3
(c) 2 + 2 3
(d) 2 2 + 3
Ans. (a)
Solution: Let the three positive numbers in the increasing G.P. be
a
, a, ar where r > 1.
r
a
We are given, , 3a, ar are in A.P.
r
\
a
+ ar = 2(3a)
r
fi
fi
As
r2 – 6r + 1 = 0
r=3± 2 2
r > 1, r = 3 + 2 2
Example 21: Suppose m arithmetic means are inserted
between 1 and 31. If the ratio of the second mean to the mth
mean is 1 : 4, then m is equal to
(a) 7
(b) 9
(c) 11
(d) 15
Ans. (b)
Complete Mathematics—JEE Main
8.8
Solution: Let m arithmetic means between 1 and 31 be
A1, A2, . . ., Am, then 1, A1, A2 . . . Am, 31 is an A.P. Let d be
the common difference of this A.P. Now,
31 = (m + 2)th term of the A.P.
= 1 + (m + 2 – 1)d
30
fi
d=
.
m +1
Also,
A2 = 1 + 2d, Am = 1 + md = 31 – d.
1
A2
Now,
=
fi 4A2 = Am
4
Am
fi
=
=
=
(1 a ) (r 2 + r + 1 + r -1 + r -2 )
m
S=
(c)
+ a = 35
n + (n + 1) x - x
(1 - x )
(n + 1) x - x
n +1
(1 - x )
n +1
(b)
2
-n
2
(d)
n - (n + 1) x + x
= (1) (1) +
n
n
=
r =1
Ê 1 - xr ˆ
 ÁË 1 - x ˜¯
r =1
=
1
n
1- x 1- x
1
1
n - (n + 1) x - x
(1 - x )2
n
 xr
r =1
1 - xr
1- x
1
m(m + 3)
4
1
(d)
m(m + 6)
4
(b)
1
1
(1 + 2) + (1 + 2 + 3)
2
3
1
(1 + 2 + 3 + 4)
4
1
(1 + 2 + 3 + . . . + m)
+...+
m
+
m
=
n +1
Â
k =1
(1 - x )2
Solution: tr = 1 + x + x2 + . . . xr – 1 =
 tr
1
1 1
1
S = Ê1 + + + . . . + ˆ
Ë
m¯
2 3
1 1
1
+ 2Ê + + . . . + ˆ
Ë2 3
m¯
1 1
1
+ 3Ê + + . . . + ˆ
Ë3 4
m¯
+...
+ (m – 1) ÊÁ 1 + 1 ˆ˜ + m Ê 1 ˆ
Ë m¯
Ë m - 1 m¯
= 49
Ans. (b)
Now,
1
Solution:
a = – 7 is not possible
a = 7 and a/r2 = 28.
a
7
Now, fifth term = ar2 = a Ê ˆ =
Ë 28 ¯ 4
Example 23: If |x| < 1, and rth term of a series is 1 + x
+ x2 + . . . + xr – 1, then sum to n terms of the series is
(a)
[n - (n + 1) x + x n+1 ]
 k ÊÁË k + k + 1 + k + 2 + . . . + m ˆ˜¯
1
m(m + 2)
4
1
m(m + 4)
(c)
4
Ans. (b)
a2 = 49 fi a = ± 7.
r2
Therefore
Thus,
(1 - x )2
(a)
a (r -2 + r -1 + 1 + r + r 2 )
a
1
[n - nx - x + x n+1 ]
is equal to:
Example 22: In a geometric progression the ratio of the
sum of the first 5 terms to the sum of their reciprocals is 49
and sum of the first and the third term is 35. The fifth term
of the G.P. is
(a) 7
(b) 7/2
(c) 7/4
(d) 7/8
Ans. (c)
a a
Solution: Let five terms in G.P. be 2 , , a, ar, ar2
r r
Then
Also,
(1 - x )2
k =1
30
= 3 fi m = 9.
m +1
fi
1
Example 24: Let m be a positive integer, then
4(1 + 2d) = 31 – d fi d = 3
Thus
1 x(1 - x n )
n
1- x 1- x 1- x
m
=
n +1
Â
k =1
1
(1 + 2 + . . . + k)
k
1 k (k + 1)
1
◊
=
2
k
2
m
 (k + 1)
k =1
m
=
[2+ (m + 1)]
4
1
=
m(m + 3)
4
È 1 3n ˘
Example 25: Let f(n) = Í +
n , where
Î 5 100 ˙˚
[x] denotes the greatest integer less than or equal to x. Then
61
 f ( n)
n =1
Progressions
(a) 2013
(c) 1947
Ans. (d)
(b) 1869
(d) 1661
3n
< 0.8 or n £ 26, then
100
È 1 3n ˘
ÍÎ 5 + 100 ˙˚ = 0.
3n
0.8 £
< 1.8, or 27 £ n £ 59
100
È 1 3n ˘
+
=1
ÎÍ 5 100 ˚˙
Solution: If
If
then
61
Thus,
Â
59
f ( n) =
n =1
Solution: We have
2n + 10 = 2 ¥ 22 + 3 ¥ 23 + 4 ¥ 24 + º + n ¥ 2n
2(2n + 10) = 2 ¥ 23 + 3 ¥ 24 + º + (n – 1) ¥ 2n +
n ¥ 2n + 1
fi
Subtracting, we get
– 2n + 10 = 2◊22 + 23 + 2 4 + º + 2n – n◊2n + 1
Â
n = 27
Â
2n
210 = 2n – 2
fi
n = 60
1
(33) (27 + 59) + 2(60 + 61)
2
= 1661
Example 26: Sum of the series S = 12 – 22 + 32 – 42 +
º – 20082 + 20092 is
(a) 2019045
(b) 1005004
(c) 2000506
(d) none of these
Ans. (a)
Solution: We can write S as
S = (1 – 2) (1 + 2) + (3 – 4) (3 + 4) + º + (2007 –
2008) (2007 + 2008) + 20092
= – [1 + 2 + 3 + 4 + º + 2008] + 20092
1
(2008) (2009) + 20092 = 2019045
2
(
)
8 2 n- 2 - 1
– n ¥ 2n + 1
2 -1
= 2 n + 1 – (n)2 n +1
=8+
61
n+
=
fi
n = 513
Example 29: Sum of the series S = 1 +
1
(1 + 2) +
2
1
1
(1 + 2 + 3) +
(1 + 2 + 3 + 4) + º upto 20 terms is
3
4
(a) 110
(b) 111
(c) 115
(d) 116
Ans. (c)
Solution: Let tk denote the kth term of the series. Then
1
1 k ( k + 1)
k +1
(1 + 2 + 3 + … + k) =
=
.
tk =
k
k
2
2
Thus,
S=
1
10
[2 + 3 + 4 + … + 21] =
(2 + 21) = 115
2
2
Example 27: The value of n for which 704 +
Example 30: If 12 + 22 + 32 + º + 20092 =
(2009) (4019) (335) and (1) (2009) + (2) (2008) + (3) (2007)
+ º + (2009) (1) = (2009) (335) (x), then x equals
(a) 2011
(b) 2009
(c) 2008
(d) 2007
Ans. (a)
Solution: According to the given condition
Solution: Let
S = (1) (2009) + (2) (2008) + (3) (2007) +
… + (2009)(1)
and
K = 12 + 22 + 32 + … + 20092
Adding, we get
S + K = (2010) [1 + 2 + 3 + … + 2009]
fi (2009) (335) (x) + (2009) (4019) (335)
= (2010) (1005) (2009)
fi
x = 2011
=–
1
(704) +
2
1
1
1
(704) + º upto n terms = 1984 –
(1984) +
(1984)
4
2
4
– º upto n terms is
(a) 5
(b) 3
(c) 4
(d) 10
Ans. (a)
{ }
(704) (2) 1 fi
Example 28: The positive integer n for which 2×22 +
3×23 + 4×24 + … + n×2n = 2n +10 is
(a) 510
(b) 511
(c) 512
(d) 513
Ans. (d)
È 1 3n ˘
= 2.
n = 60, 61 Í +
Î 5 100 ˙˚
For
8.9
1
2n
=
(1984 ) (2 ) Ï
Ô
2112
3
(-1)n ¸Ô
Ì1 - n ˝
2 ˛Ô
ÓÔ
n
(-1)
- 1984 n
2n
2
n
If n is odd, we get 2 = 32 or n = 5
If n is even, we get 128 = 128/2 n fi n = 0.
128 =
Example 31: If x > 0, and log2 x + log 2
log 2
( 4 x ) + log2 ( 8 x ) +
log2
(16 x ) + º = 4
( x)
+
8.10
Complete Mathematics—JEE Main
then x equals
(a) 2
(c) 4
Ans. (c)
Example 34: Suppose a, b, c are in A.P. and a2, b2, c2
are in G.P. If a < b < c and a + b + c = 3/2, then the value
of a is
(b) 3
(d) 5
(a)
Solution: We can write the given equation as
Ê 1+
log2 Á x
Ë
fi
log2(x2) = 4
1 1 1 1
+ + + + ˆ
2 4 8 16
˜ =4
¯
fi
x2 = 24
fi
(c)
x=4
Example 32: If (1 + 3 + 5 + º + p) + (1 +
3 + 5 + º + q) = (1 + 3 + 5 + º + r) where each set
of parentheses contains the sum of consecutive odd integers
as shown, the smallest possible value of p + q + r, (where
p > 6) is
(a) 12
(b) 21
(c) 45
(d) 54
Ans. (b)
Solution: We know that
1 + 3 + 5 + … + (2k – 1) = k2
Thus, the given equation can be written as
2
2
2
Ê q + 1ˆ Ê r + 1ˆ
Ê p + 1ˆ
ÁË
˜¯ + ÁË
˜¯ = ÁË
˜¯
2
2
2
fi (p + 1)2 + (q + 1)2 = (r + 1)2
Therefore, (p + 1, q + 1, r + 1) forms a Pythagorean triplet.
As p > 6, p + 1 >7.
The first Pythagorean triplet containing a number > 7 is
(6, 8, 10).
\ We may take p + 1 = 8, q + 1 = 6, r + 1 = 10
fi
p + q + r = 21.
Example 33: Let a1, a2, º, a10 be in A.P. and h1, h2, º,
h10 be in H.P. If a1 = h1 = 2 and a10 = h10 = 3, then a4 h7 is
(a) 2
(b) 3
(c) 5
(d) 6
Ans. (d)
Solution: Let d be the common difference of the A.P.,
then
3 = a10 = 2 + 9d
fi
d = 1/9
\
a4 = 2 + 3d = 7/3
Next, let D be the common difference of the A.P.
1 1
1
, , ... ,
Then
h1 h2
h10
\
Hence,
1 1 1
=
= + 9D
3 h10 2
fi
1
7
1
=
+ 6D =
h7
18
h1
fi
a4 h7 = (7/3) (18/7) = 6.
D= h7 =
1
54
18
7
1
(b)
2 2
1 1
2
3
(d)
1
2 3
1 1
2
2
Ans. (d)
Solution: We have a + c = 2b fi 3b = a + b + c = 3/2
fi b = 1/2
Thus, a + c = 1 and a2 c2 = b4 = 1/16
fi
a2(1 – a)2 = 1/16 fi a(1 – a) = ± 1/4
fi 4a2 – 4a ± 1 = 0 fi (2a – 1)2 = 0 or (2a – 1)2 = 2
fi
a=
1
1 1
or a = ±
2
2
2
As a < b < c and b =
1
1 1
, we get a = 2
2
2
Example 35: Let S1, S2, º be squares such that for
each n ≥ 1, the length of a side of Sn equals the length of a
diagonal of Sn +1. If the length of a side of S1 is 10 cm, then
the smallest value of n for which Area (Sn) < 1 is
(a) 7
(b) 8
(c) 9
(d) 10
Ans. (b)
Solution: Let an denote the length of a side of Sn. We
are given
Length of a side of Sn = Length of a diagonal of Sn+1
an +1
1
fi
an = 2 an + 1 fi
=
.
an
2
Thus, a1, a2, a3, º is a G.P. with first term 10 and common
ratio 1/ 2 . Therefore,
(
an = 10 1 / 2
)
n -1
Area (Sn) = a 2n < 1
Also,
fi [10 (1/ 2 )n – 1]2 < 1 fi 100 < 2n – 1
fi smallest possible value of n is 8.
Example 36: Let Tr be the rth term of an AP, for r = 1,
1
2, º. If for some positive integers m and n, we have Tm =
n
1
, the Tm + n equals
and Tn =
m
1
mn
1
(c)
m
Ans. (c)
(a)
(b)
1 1
+
m n
(d) 0
Progressions
Solution: Let a be the first term and d be the common difference of the given A.P. Then according to the
hypothesis,
1
1
Tm – Tn =
–
m
n
fi
fi
(m – n)d =
m–n
mn
fi
1
.
mn
d=
Example 37: Consider an infinite geometric series with
the first term a and common ratio r. If its sum is 4 and the
second term is 3/4, then
(a) a = 4/7, r = 3/7
(b) a = 2, r = 3/8
(c) a = 3/2, r = 1/2
(d) a = 3, r = 1/4
Ans. (d)
Solution: We have
a
3
= 4 and ar =
1- r
4
Thus,
a
=4
1 - 3 / 4a
fi
fi
r=
3
.
4a
a2 – 4a + 3 = 0 fi a = 1, 3
When a = 1, r = 3/4 and when a = 3, r = 1/4.
Example 38: If the sum of first 2n terms of the A.P. 2,
5, 8, º, is equal to the sum of first n terms of the A.P. 57,
59, 61, º, then n equals
(a) 10
(b) 12
(c) 11
(d) 13
Ans. (c)
Solution: Sum to 2n terms of 2, 5, 8, º is
2n
[2(2) + (2n – 1) (3)] = n(6n + 1)
2
and sum to n terms of 57, 59, 61, º, is
n
[2 (57) + (n –1) (2)] = n(56 + n)
2
According to given condition
n(6n + 1) = n(56 + n) fi 6n + 1 = 56 + n fi n = 11
Example 39: For a positive integer n, let
1 1 1
1
. Then
a(n) = 1 + + + + +
n
2 3 4
2 -1
( )
(a) a (100) < 100
(c) a (200) < 100
Ans. (a)
2 4 8
2 n -1
+ + + + n -1 = n
2 4 8
2
Thus, a (100) < 100.
Example 40: If x > 1, y > 1, z > 1 are in G.P., then
1
1
1
,
,
are in
1 + ln x 1 + ln y 1 + ln z
(a) A.P.
(c) H.P.
Ans. (c)
1
1
=
mn
m
Tm + n – Tm = (m + n – m)
<1+
8.11
(b) G.P.
(d) none of these
Solution: As x, y, z are in G.P. fi y2 = xz
fi
ln(y2) = ln(xz)
fi
2ln(y) = ln(x) + ln(z)
fi
2(1 + ln(y)) = (1 + ln (x)) + (1 + ln (z))
fi
1+ ln(x), 1 + ln(y), 1 + ln(z) are in A.P.
fi
1
1
1
,
,
are in H.P.
1 + ln ( x ) 1 + ln ( y ) 1 + ln ( z )
Example 41: Let x be the arithmetic mean and y, z be
the two geometric means between any two positive numbers.
y3 + z 3
Then value of
is
x yz
(a) 2
(b) 3
(c)
1
2
3
2
(d)
Ans. (a)
Solution: Let two positive numbers be a and b. Then
x = (a + b)/2. Also, a, y, z, b are in G.P. If r is the common
ratio of this G.P., then b = ar3 fi r = (b/a)1/3.
We have
a 1 + r3
y3 + z 3
a 3r 3 + a 3r 6
a+b
=
=
=
=2
2
x yz
x
(a + b) /2
x ( ar ) ar
(
)
( )
Example 42: The sum to n terms of the series
1
1+ 2
1+ 2 + 3
+
+
+ º is
13 13 + 23 13 + 23 + 33
n
n
(b)
(a)
n +1
2 ( n + 1)
(c)
2n
n +1
(d)
2
n ( n + 1)
Ans. (c)
(b) a (100) > 100
(d) none of these
Solution: If tn denotes the nth term of the series, then
Solution: We have
1ˆ
Ê 1 1ˆ Ê 1
a(n) = 1 + Á + ˜ + Á + + ˜
Ë 2 3¯ Ë 4
7¯
tn =
Ê 1
1ˆ
1
1 ˆ
Ê1
+ Á + + ˜ + + Á
+
+
+
˜
n -1
Ë8
15 ¯
2 n -1 + 1
2 n - 1¯
Ë2
( )
=
1 + 2 + 3 + + n
13 + 23 + 33 + + n3
1
n ( n + 1)
= 2
1 2
2
n ( n + 1)
4
2
= 2 È1 - 1 ˘
ÍÎ n n + 1 ˙˚
n ( n + 1)
8.12
Complete Mathematics—JEE Main
n
Â
fi
k =1
{
Ê 1 ˆ Ê 1 1ˆ
tk = 2 Á 1 - ˜ + Á - ˜ + +
Ë 2 ¯ Ë 2 3¯
1 ˆ
Ê1
˜
ÁË n n + 1¯
2n
1 ˆ
Ê
= 2 Á1 ˜¯ =
Ë
n +1
n +1
Example 43: Let
then (a2022)2022 equals
(a) – 1
(c) 0
Ans. (b)
}
n
Â
r =1
Â
r =1
(b) f (2n) – 7f (n)
(c) f (2n – 1) – 8 f (n) (d) none of these
Ans. (a)
Solution: We have
2n
n
Â
r =1
(2r – 1)4 =
Â
r =1
n
r4 –
Â
1
and
1 - a1
(2r – 1)4 is
a3 =
equal to
(a) f (2n) – 16 f (n)
(b) 1
(d) none of these
Solution: We have a2 =
n
r4 = f (n), then
1
for n ≥ 1 and a3 = a1,
1 - an
Example 46: If an + 1 =
(2r)4 = f (2n) – 16f (n)
r =1
Example 44: The sum of first n terms of the series 12 +
2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 + … is n(n + 1)2/2 when n
is even. When n is odd the sum of the series is
(b) n2(n + 1)/2
(a) n2(3n + 1)/4
3
(d) none of these
(c) n (n – 1)/2
Ans. (b)
Solution: Let n = 2m, then
S2m = 12 + 2×22 + 32 + 2×42 + … +
(2m – 1)2 + 2(2m)2
2
= 2m(2m + 1) /2 = m(2m + 1)2
When n = 2m – 1,
S2m – 1 = 12 + 2×22 + 32 + 2×42 + … +
(2m – 1)2
= S2m – 2(2m)2 = m(2m + 1)2 – 2(2m)2
= m [4m2 + 4m + 1 – 8m] = m(2m – 1)2
= n2 (n + 1)/2
Example 45: If the first and (2n+ 1)th terms of an A.P.;
G.P. and H.P. are equal and their (n + 1)th terms are a, b and
c respectively, then
(a) a > b > c
(b) ac = b2
(c) a + b = c
(d) none of these
Ans. (b)
Solution: Let A be the first term, D be the common difference and B be the (2n + 1)th term of the A.P., then
B- A
B = A + 2nD fi D =
.
2n
A+ B
.
Also,
a = A + (n + 1 – 1)D =
2
2AB
Similarly,
b = AB and c =
.
A+ B
Clearly, we have
b2 = ac
1 - a1
1
1
=
=
- a1
1 - a2
1 - 1/ (1 - a1 )
Since a3 = a1, we get – a 21 = 1 – a1
fi a 12 – a1 + 1 = 0 fi a1 = – w or – w2
1 - a3 1 - a1
1
1
We have
a5 =
=
=
=
- a3
- a1
1 - a4 1 - 1/ (1 - a3 )
= a3 = a1
Continuing in this way we obtain
a1 = a3= a5 = a7 = a9 = a11 = º = a2022
Thus, (a2022)2022 = (– w)2022 or (– w2)2022 = 1
Example 47: If f (x + y) = f (xy) " x, y e R, and f (2013)
= 2013, then f (– 2013) equals
(a) 2013
(b) 0
(c) – 2013
(d) none of these
Ans. (a)
Solution: f (2013) = f (2013 + 0) = f [(2013 (0)] = f (0)
fi
f(0) = 2013
Also,
f (– 2013) = f (– 2013 + 0) = f(0) = 2013.
It can also be seen that f is a constant function.
n
Example 48: If
Â
tr =
r =1
n
1
Ât
1
n(n + 1) (n + 2), the value
12
is
r =1 r
(a)
2n
n +1
(b)
n -1
(n + 1)!
(c)
4n
n +1
(d)
3n
n+2
Ans. (c)
r -1
r
Solution: We have tr =
Â
k =1
tk –
Â
tk
k =1
1
1
r(r + 1) (r + 2) –
(r – 1) (r) (r + 1)
12
12
1
r(r + 1)
=
4
=
Progressions
n
Â
fi
Ans. (a)
Solution: We have
1 ˆ
1
4
Ê1
=
= 4Á Ë r r + 1˜¯
tr
r (r + 1)
Now,
r =1
n
1
1 ˆ
1 ˆ
4n
Ê1
Ê
= 4 Á ˜¯ = 4 ÁË 1 ˜¯ =
Ë
1
tr
r
r
+
n
+
1
n +1
r =1
15
+ is
16
n
2
n +1
(b)
n
1
2 n +1
(d) n – 1 +
2n
Solution: We have tr = 1 –
n
n
Â
tr =
r =1
Â
r =1
1
2r
(
) (
Â
2n
sin 2n f and z =
Â
n= 0
(
•
Â
•
2n
cos f,
\
n= 0
cos2n f sin2n f, then
1
2
1 - cos f
=
1
sin 2 f
1
cos2 f
Also,
Also,
z=
fi
1
2
2
1 - cos f sin f
=
1
1
1xy
=
xy
xy - 1
xyz – z = xy or
xyz = xy + z
1 1
+ = 1 fi x + y = xy
x y
xyz = x + y + z
But
\
1
n(n + 1)
2
(b)
1
n(n + 2)
2
(c)
1
n(n + 3)
2
(d)
1
n(n + 5)
2
Example 51: The sum to n terms of the series
1
1
1
+
+
+ is
7 + 10
10 + 13
13 + 16
4 + 3n - 2
3
(a)
1
3
(
7 + 3n - 7
(c)
1
3
(
10 + 3n - 10 (d) none of these
)
(b)
)
tn = Sn – Sn–1 " n ≥ 2
1
1
n(2n2 + 9n + 13) –
(n – 1) {(2(n – 1)2
6
6
+ 9(n – 1) + 13}
1
[2(n3 – (n – 1)3) + 9(n2 – (n – 1)2)
=
6
+ 13(n – n + 1)]
1
[6n2 – 6n + 2 + 9(2n – 1) + 13]
=
6
1
=
(6n2 + 12n + 6)
6
= (n + 1)2
t1 = S1 = 4 = (1+1)2
tn =
(b) xyz = xz + z
(d) xyz = yz + z
x=
(a)
Solution: We have
n= 0
Solution: We have
)
Ans. (c)
2n
Â
)
tr equals
)
1
(a) xyz = xz + y
(c) xyz = x + y + z
Ans. (c)
and y =
(
r =1
for r ≥ 1.
p
Example 50: If 0 < f < , and x =
2
•
)
n
1
(1 / 2) 1 - 1/ 2n
1ˆ
Ê
1
=
n
–
ÁË
˜
1 - 1/ 2
2r ¯
=n–1+
7 + 3r - 4 + 3r
3
4 + 3r + 7 + 3r
Thus, sum to n terms is equal to
1
È 10 - 7 + 13 - 10 + + 7 + 3n - 4 + 3n ˘˚
3Î
1
7 + 3n - 7
=
3
n
1
Example 52: If Sn = Â tr =
n(2n2 + 9n + 13), then
6
r =1
Ans. (d)
y=
=
(
(c) n + 1 –
Thus,
1
tr =
1 3 7
+ + +
2 4 8
Example 49: Sum to n term of the series
(a)
8.13
n
n
\ Â tr = Â (r + 1) =
r =1
r =1
1
1
(n + 1) (n + 2) – 1 =
n(n + 3)
2
2
Example 53: The interior angles of a convex polygon
are in A.P. If the smallest angle is 100∞ and the common
difference is 4∞, then the number of sides is
(a) 5
(b) 7
(c) 36
(d) 44
Ans. (a)
Solution: We know that the sum of the interior angles
of a convex polygon of n sides is (n – 2) (180∞). We are given
a = 100 and d = 4.
n
\
[2(100) + (n – 1) (4)] = (n – 2) (180)
2
fi
n = 5, 36
fi n2 – 41n + 180 = 0
If n = 36, then interior angles of the polygon are 100∞, 104∞,
º, 176∞, 180∞, 184∞, º, 244º.
8.14
Complete Mathematics—JEE Main
But no interior angle can be 180∞. Therefore, n = 5
Example 54: If a and x are positive integers such that
x < a and a - x , x , a + x are in A.P., then least possible value of a is
(a) 5
(b) 7
(c) 11
(d) none of these
Ans. (a)
Solution: We have 2 x = a + x + a - x
fi 2 x =
2x
a+x - a-x
a+x - a-x =
fi
a
4
=
a+d
5
sin (90∞ – q) = 3/5
We have
and
sin q =
90∞ - q
a+d
a
q
x
5
x
4
As a and x are positive integers, least possible value of x is
4 and that of a is 5.
a-d
Fig. 8.1
\ 3 x = 2 a + x fi 4(a + x) = 9x fi a =
1
Example 55: If
+
1
1
+
+ upto • =
12 22 32
1 1
1
value of 2 + 2 + 2 + up to • is
1
3
5
2
p
p2
(a)
(b)
4
6
p2
(c)
8
Ans. (c)
p2
, then
6
p2
(d)
12
Solution: We have
=
Solution: We have
=
1
12
+
1
2
1
1
22
+
1
+
2
3
+
1
32
1
52
+
1
42
+ upto •
+
1
52
+
1
62
=
5 -1
,
2
5 +1
(d)
2
= tan2 x
a = exp {(sin2 x + sin4 x + sin6 x + º upto
= exp {tan2 x ln 2} = exp {ln 2
}
=2
As a satisfies the equation x2 – 17x + 16 = 0 we get
a = 1 or a = 16
2
Since 0 < x < p/2, tan2 x > 0 fi a = 2tan x > 1. Therefore,
2tan
Thus,
2 / 3 , 1/ 3
3 -1
,
2
tan2 x
tan2 x
Example 56: If the length of sides of a right triangle are
in A.P., then the sines of the acute angle are
(b)
1 - sin 2 x
Therefore,
upto •
p2 1 Êp2 ˆ
p2
- Á ˜ =
6 4Ë 6 ¯
8
(a) 3/5, 4/5
sin 2 x
sin2 x + sin4 x + sin6 x + º
•) ln 2}
1 È
1
1
˘
– 2 Í1 + 2 + 2 + ˙
˚
2 Î 2
3
(c)
Example 57: If exp {(sin2 x + sin4 x + sin6 x + º upto
•) ln 2} satisfies the equation x2 – 17x + 16 = 0 then value
2 cos x
of
(0 < x < p/2) is
sin x + 2 cos x
1
3
(a)
(b)
2
2
5
(c)
(d) none of these
2
Ans. (a)
3 +1
2
Ans. (a)
Solution: Let the sides of the triangle be a – d, a, a + d
where a > d > 0.
By the Pythagorean theorem, we have
(a + d)2 = (a – d)2 + a2
fi
4ad = a2 fi d = a/4
2x
= 16 = 24 fi tan2 x = 4 fi tan x = 2
[ tan x > 0]
2 cos x
2
2
1
=
=
=
sin x + 2 cos x
tan x + 2
2+2
2
Example 58: Let f(x) be a polynomial function of second degree. If f(1) = f(–1) and a, b, c are in A.P., then f ¢(a),
f ¢(b), f ¢(c) are in
(a) G.P.
(b) H.P.
(c) A.G.P.
(d) A.P.
Ans. (d)
Solution: Let f(x) = Ax2 + Bx + C.
fi
\
f(1) = f(– 1) fi A + B + C fi A – B + C
2B = 0 fi B = 0.
f(x) = Ax2 + C fi f ¢(x) = 2Ax
Progressions
As a, b, c are in A.P., 2Aa, 2Ab, 2Ac are in A.P., i.e., f ¢(a),
f ¢(b), f ¢(c) are in A.P.
Example 59: If exp {(tan2 x – tan4 x + tan6 x – tan8 x + º)
loge 16}, 0 < x < p/4, satisfies the quadratic equation x2 –
3x + 2 = 0, then value of cos2 x + cos4 x is
(a) 4/5
(b) 21/16
(c) 17/11
(d) 19/31
Ans. (b)
Solution: We have
tan2 x – tan4 x + tan6 x – tan8 x + º
=
tan 2 x
(
1 - - tan 2 x
)
tan 2 x
=
sec 2 x
= sin2 x
Therefore
y = exp {(tan2 x – tan4 x + tan6 x – tan8 x + º) log e 16}
= exp {(sin2 x) log e16}
2
2
= exp {loge (16 sin x)} = 16 sin x
As y satisfies
x2 – 3x + 2 = 0, we get y = 1 or y = 2
16sin
fi
2
x
sin
= 1 or 16
2
x
=2
Since 0 < x < p/4, 0 < sin x < 1/ 2
\
16sin
2
16sin
2
Thus,
fi
x
= 1 is not possible.
x
= 2 fi sin2 x = 1/4
0 < sin2 x < 1/2
(a) 41
(c) 43
Ans. (a)
value of
1 1 1 1 1
p
+ - + - + = , then
3 5 7 9 11
4
1
1
1
+
+
+ is
1 ◊ 3 5 ◊ 7 9 ◊ 11
p
(a)
8
p
(c)
4
Ans. (a)
p
(b)
6
p
(d)
34
Solution: We have
p
Ê 1ˆ Ê 1 1ˆ Ê 1 1 ˆ
= Á1 - ˜ + Á - ˜ + Á - ˜ + Ë 3 ¯ Ë 5 7 ¯ Ë 9 11¯
4
2
2
2
+
+
=
+
1 ◊ 3 5 ◊ 7 9.11
fi
1
1
1
p
+
+
+ =
1 ◊ 3 5 ◊ 7 9 ◊ 11
8
Example 61: For x e R, let [x] denote the greatest integer £ x. Largest natural number n for which
p˘ È 2
p˘
p˘
Èp ˘ È 1
È n
+ ˙…+Í
+ ˙ < 43,
E= Í ˙+Í
+ ˙+Í
Î 2 ˚ Î100 2 ˚ Î100 2 ˚
Î100 2 ˚
is
(b) 42
(d) 97
Solution: Since 3.14 < p < 3.142, 1.57 < p/2 < 1.571
n ˘
Èp
Thus, Í +
= 1 for n = 0, 1, 2, º, 42
Î 2 100 ˙˚
Thus, largest natural number n for which E < 43 is 41.
Example 62: If the ratio of sums to n terms of two A.P.’s
is (5n + 7) : (3n + 2), then the ratio of their 17th terms is
(a) 172 : 99
(b) 172 : 101
(c) 175 : 99
(d) 175 : 101
Ans. (b)
Solution: Let two A.P.’s be
a, a + d, a + 2d, a + 3d, º
and
A, A + D, A + 2D, A + 3D, º
According to the given condition,
n
[2a + (n - 1) d ] 5n + 7
2
=
n
[2 A + (n - 1) D ] 3n + 2
2
Putting
Thus, cos2 x + cos4 x = (1 – sin2 x) + (1 – sin2 x)2 = 21/16
Example 60: If 1 –
8.15
n -1
d
5n + 7
2
=
n -1
3n + 2
A+
D
2
a+
fi
n -1
= 16 or n = 33, we get
2
a + 16 d
172
5 (33) + 7
=
=
A + 16 D
101
3 (33) + 2
Example 63: If H1, H2, º, Hn are n harmonic means
between a and b( a ), then value of
equal to
(a) n + 1
(c) 2n
Ans. (c)
H1 + a H n + b
+
is
H1 - a H n - b
(b) n – 1
(d) 2n + 3
Solution: As a, H1, H2, º, Hn b are in H.P.,
1 1 1
1 1
, ,
, are in A.P.
a H1 H 2
Hn b
Let d be the common difference of this A.P., then
1
1
=
+ (n + 1)d
b
a
1
1
–
= (n – 1)d
and
H n H1
Now,
and
\
H1 + a
1 / a + 1 / H1 1 / a + 1 / H1
=
=
H1 – a
1 / a - 1 / H1
-d
Hn + b
1 / b + 1 / Hn 1 / b + 1 / Hn
=
=
Hn – b
1 / b - 1 / Hn
d
H1 + a H n + b
1 / a + 1 / H1 1 / b + 1 / H n
+
+
=
H1 - a H n - b
-d
d
8.16
Complete Mathematics—JEE Main
=
= – 2nad – n(2n – 1)d2 + a2 + 4 n(ad) + 4n2d2
= a2 + 2nad + n(2n + 1)d2 = (a + nd)2 + n(n + 1)d 2
1 ÈÊ 1 1 ˆ Ê 1
1 ˆ˘
= 2n
Á – ˜+
d ÍÎË b a ¯ ÁË H n H1 ˜¯ ˙˚
Example 64: If three positive real numbers a, b, c are
in A.P. such that abc = 4, then the minimum possible value
of b is
(b) 22/3
(a) 23/2
1/3
(c) 2
(d) 25/2
Ans. (b)
Example 67: Value of y = (0.36)
is
(a) 0.9
(c) 0.6
Ans. (c)
b3 = 4 + bd2 ≥ 4
[
b > 0, d 2 ≥ 0]
b ≥ 22/3
fi
Thus, minimum possible value of b is 22/3, that is the case
when d = 0.
\
fi
2
x + |cos3 x | + upto •
= 34 are given by
(a)
p 2p
,
3 3
(b)
p 3p
,
3 4
(c)
p 3p
,
4 4
(d)
p p
,
3 4
Solution: Since 0 < x < p, – 1 < cos x < 1 fi
0 £ |cos x| < 1. We can write the given expression as
91/(1 – |cos x|) = 81 = 92
1
=2
1 - cos x
fi
|cos x| = 1/2
fi
fi
1 – |cos x| =
fi
cos x = ± 1/2
fi
1
2
Example 66: The sum upto (2n + 1) terms of the series
a2 – (a + d)2 + (a + 2d)2 – (a + 3d)2 + º is
a2 + 3nd2
a2 + 2nad + n(n – 1)d2
a2 + 3nad + n(n – 1)d2
(a + nd)2 + n(n + 1)d2
fi
fi
Solution: We can write the sum upto (2n + 1) terms as
[a + (a + d)] (– d) + [(a + 2d) + (a + 3d)] (– d) + º
[(a + (2n – 2) d) + (a + (2n – 1)d] (– d ) + (a + 2nd)2
1 - xn
- nx n
1- x
È Ê
1ˆ n ˘
1ˆ n
Ê 1ˆ
Ê
ÁË - ˜¯ S = (– n) Í1 - ÁË 1 + ˜¯ ˙ - n ÁË 1 + ˜¯ = – n
n ˚
n
n
Î
2
S=n
Example 69: Suppose a, b, c are positive real numbers.
If a, b, c are in A.P. and a2, b2, c2 are in H.P., then
(a) a = b = c
(b) 2b = 3a + c
2
(d) none of these
(c) b = ac/8
Ans. (a)
2a 2 c2
Solution: We have 2b = a + c and b2 = 2
a + c2
= (– d) [a + (a + d) + (a + 2d) + …
+ a + (2n – 1)d] +(a + 2nd)
2n
= (– d)
{a + a + (2n – 1) d} + (a + 2nd)2
2
- log 2
1
log (0.36) = log (0.36) = log 0.6
- log 4
2
y = 0.6
=
2
log (1/ 2 )
log (0.36)
log (0.25)
Solution: Let x = 1 + 1/n. Then
S = 1 + 2x + 3x2 … + nxn –1
fi
xS = x + 2x2 + …+ (n – 1)xn – 1 + nxn
Subtracting, we get
(1 – x)S = 1 + x + x2 + … + xn – 1 – nxn
x = p/3 or 2p/3
(a)
(b)
(c)
(d)
Ans. (d)
log y =
Example 68: Sum to n terms of the series
1ˆ
1ˆ 2
Ê
Ê
S = 1 + 2 Á 1 + ˜ + 3 Á 1 + ˜ + is given by
Ë
Ë
n¯
n¯
2
(b) (n + 1)2
(a) n
(c) n(n + 1)
(d) none of these
Ans. (a)
Ans. (a)
fi
(1 / 2 )
y = (0.36)log0.25
=
Example 65: For 0 < x < p the values of x which satisfy
the relation 91 + |cos x| + cos
(b) 0.8
(d) 0.25
Solution: We have
1/ 3
1
1 1
1
S= +
=
+ 3 + =
2
2
1
1
/
3
3 3
3
Solution: Let d be the common difference of the A.P.,
then
4 = abc = (b – d)b (b + d) = b(b2 – d2)
fi
1 1 1
log0.25 ÊÁ + 2 + 3 +º upto •ˆ˜
Ë3 3 3
¯
\
fi
2
2a 2 c2
Ê a + cˆ
=
fi (a + c)2 (a2 + c2) = 8a2c2
ÁË
˜
2 ¯
a2 + c2
(a2 + c2)2 + 2ac(a2 + c2) – 8a2c2 = 0
Progressions
fi
(a2 + c2 + 4ac) (a2 + c2 – 2ac) = 0
fi
[(a + c)2 + 2ac] (a – c)2 = 0
fi
a – c = 0 or (a + c)2 + 2ac = 0
As (a + c)2+ 2ac > 0, we get a = c fi a = b = c
•
Example 70: If 0 < q, f < p/2 and x =
Â
Ans. (b)
Solution: If tr denotes the rth term of the series, then
1
x
1
x tr =
=
(1 + rx )(1 + (r + 1) x ) 1 + rx 1 + (r + 1) x
sin2n f,
n
•
2n
n
n= 0
(a) xyz + 1 = yz – zx (b) xyz – 1 = yz + zx
(c) xyz – xy = yz – zx (d) xyz + 1 = yz + zx
Ans. (c)
1
1
Solution: We have x =
=
,
2
1 - sin f cos2 f
1
1
=
,
y=
2
sin 2 q
1 - cos q
1 1
Also, cos (q + f) cos (q – f) = cos2f – sin2 q = x y
As, 0 < cos2 f < 1, 0 < sin2 q < 1, –1< cos2 f – sin2 q < 1
1 1
fi
–1 < - <1
x y
•
n
Ê 1 1ˆ
ÁË x - y ˜¯ =
1
xy
=
xy - y + x
Ê 1 1ˆ
n=0
1- Á - ˜
Ë x y¯
fi
z (xy – y + x) = xy or xyz – xy = yz – xz
b b
b
Example 71: If a, b, c are in H.P. then a – , , c 2 2
2
are in
(a) A.P.
(b) G.P.
(c) A.G.P.
(d) H.P.
Ans. (b)
Thus,
z=
Solution:
Ê
Ëa –
bˆÊ
c–
2¯Ë
Â
b=
nx
(a)
(1 + x )(1 + nx )
n
(b)
(1 + x ) [1 + (n + 1) x ]
x
(d) none of these
(1 + x )(1 + (n - 1) x )
˘
1
1
1 + x 1 + (n + 1) x
=
nx
(1 + x )(1 + (n + 1) x )
=
n
(1 + x ) [1 + (n + 1) x ]
Example 73: If a1, a2, ... an are in H.P. then the expression a1a2 + a2a3 + ... + an – 1 an is equal to
(b) n(a1 – an)
(a) (n – 1)a1 an
(d) na1 an
(c) (n – 1) (a1 – an)
Ans. (a)
Solution: a1, a2, ... an are in H.P.
1 1
1
,
, ...
are in A.P.
a1 a2
an
fi
1
1
1
1
1
1
–
=
=º
= d (say)
–
–
a
a
a
a
a2 a1
3
2
n
n –1
fi
fi
a 1a 2 =
Thus,
b
b2
bˆ
= ac – (a + c) +
2
4
2¯
Example 72: Sum to n terms of the series
1
1
1
+
+
+ º is
(1 + x )(1 + 2 x ) (1 + 2 x )(1 + 3 x ) (1 + 3 x )(1 + 4 x )
=
r =1
2ac
, and
a+c
b2
b2
= ac – ac +
=
4
4
(c)
 tr
1
r =1
n
fi
È 1
 ÍÎ1 + rx - 1 + (r + 1) x ˙˚
r =1
n
y = Â cos q and z = Â cos (q + f ) cos (q – f), then
n= 0
n
x  tr =
fi
n= 0
•
8.17
But
\
1
1
(a1 – a2), a2a3 = (a2 – a3), ...
d
d
1
an – 1 an = (an – 1 – an)
d
a1a2 + a2a3 + ... + an – 1an
1
= [a1 – a2 + a2 – a3 + ... + an – 1 – an]
d
1
= (a1 – an)
d
a1 – an
1
1
= + (n – 1)d fi
= (n – 1)d
an
an a1
a1
a1a2 + a2a3 + ... + an – 1 an = (n – 1)a1 an
Example 74: If the sum of the series 2 +
+ ... is finite, then
(a) |x| > 5
(c) |x| < 5/2
Ans. (a)
(b) – 5 < x < 5
(d) |x| > 5/2
Solution: We can rewrite the series as
1+1+
5 25 125
+
+
x x2 x3
5 Ê 5ˆ 2 Ê 5ˆ 3
+ Á ˜ + Á ˜ +º
Ë x¯
x Ë x¯
8.18
Complete Mathematics—JEE Main
We can sum up this series if Ω
5
Ω<1
x
¤
|x| > 5
Example 75: In a G.P. consisting of positive terms,
each term equals the sum of the next two terms. Then common ratio of the G.P. is
1
1
(1 - 5 )
5
(b)
(a)
2
2
(c)
r +1
= r [log (r + 1) – log r]
r
= (r + 1) log (r + 1) – r log r – log (r + 1)
Solution: Let tr = r log
n
fi
n
 tr
ar n = ar n + 1 + ar n + 2 " n ≥ 1
1
fi 1 = r + r 2 fi r2 + r – 1 = 0 fi r = (- 1 ± 5 )
2
As G.P. is of positive terms, r > 0. Thus, r =
1
( 5 - 1)
2
Example 76: Suppose a, b, c, are in A.P. and – 1 < a, b,
•
•
•
n= 0
n= 0
n= 0
c < 1. Let x = Â a n , y = Â b n , z = Â c n , then x, y, z are in
(a) A.P.
(c) H.P.
Ans. (c)
1
1
1
, y=
, z=
.
1- a
1- b
1- c
As a, b, c are in A.P., – a, – b, – c are in A.P. fi 1– a, 1– b,
1– c are in A.P.
1
1
1
,
,
are in H.P.
fi
1- a 1- b 1- c
x=
Example 77: If log52, log5 (2x – 5) and log5(2x – 7/2)
are in A.P., then x is equal to
(a) 7
(b) 3
(c) 4, 5
(d) 8
Ans. (b)
Solution: As log52, log5(2x – 5), log5(2x – 7/2) are in
A.P., we get
2 log5(2x – 5) = log5 2 + log5(2x – 7/2)
fi (2x – 5)2 = 2(2x – 7/2)
fi (2x )2 – 12 (2x ) + 32 = 0 fi (2x – 4) (2x – 8) = 0
fi
2 x = 2 2, 2 3
fi
x = 2, 3.
Clearly, x π 2. Therefore x = 3
n
Example 78: Sum of the series
 r log
r =1
n
(n + 1)
n!
(c) n! log (n + 1)
(a) log
r +1
is
r
log (n + 1)
n!
(d) none of these
(b)
r =1
= log
r =1
(n + 1)n +1
(n + 1)n
= log
n!
(n + 1)!
Example 79: Let A1, A2 be two arithmetic means, G1,
G2 be two geometric means, and H1, H2 be two harmonic
means between two positive numbers a and b, then values
G1 G2 H1 + H 2
is
◊
H1 H 2 A1 + A2
1
2
(c) 1
Ans. (c)
3
2
(d) 2
(a)
(b)
Solution: We have A1 + A2 = a + b,
(b) G.P.
(d) none of these
Solution: We have
n
 {(r + 1) log(r + 1) - r log r} -  log (r + 1)
= (n + 1) log (n + 1) – 1log1 – log (n + 1)!
Ans. (d)
Solution: We are given
=
r =1
1
( 5 - 1)
2
(d)
5
Ans. (a)
G1G2 = ab and
\
1 1
1
1
= + .
+
a b
H1 H 2
1
1 ˆ
G1 G2 H1 + H 2
G G
= 1 2 ÊÁ
+
◊
˜
H1 H 2 A1 + A2
A1 + A2 Ë H1 H 2 ¯
=
ab Ê 1 1 ˆ
= 1.
+
a + b Ë a b¯
Example 80: Sum of the series Sn = (n) (n) +
(n – 1) (n + 1) + (n – 2) (n + 2) + ... + 1(2n – 1) is
(a)
1
1 3
n (n + 1) (n + 2) (b)
n – n2
6
3
(c) n3
Ans. (d)
(d) none of these.
n -1
Solution: Sn =
n -1
 (n - r ) (n + r ) =  (n2 - r 2 )
r=0
r=0
1
(n – 1) n(2n – 1)
6
Example 81: Suppose for each n ΠN, 14 + 24 + 34 +
... + n4 = an5 + bn4 + cn3 + dn2 + en + f, then value of b is
(a) 1/5
(b) 1/2
(c) 1/3
(d) 1
Ans. (b)
= n3 –
Solution: We have
14 + 24 + ... + n4 = an5 + bn4 + cn3 + dn2 + en + f
and
14 + 24 + ... + n4 + (n + 1)4
(1)
Progressions
= a(n + 1)5 + b(n + 1)4 + c(n + 1)3 +
d (n + 1)2 + e(n + 1) + f
Subtracting (1) from (2), we get
(n + 1)4 = a[(n + 1)5 – n5] + b[(n + 1)4 – n4]
+ c [(n + 1)3 – n3] + d[(n + 1)2 – n2]
+ e [(n + 1) – n]
Comparing coefficients fo n4, we get
1 = 5a fi a = 1/5
Comparing coefficient fo n3, we get
4 = a (5C2) + 4b fi b = 1/2.
(2)
Example 82: Value of S = 2012 +
1Ê
1Ê
1
1Ê
1 ˆ ˆ
Á 2011 + ÁË 2010 + (2009 + ... + ÁË 2 + (1)˜¯ ...˜¯ is
3Ë
3
3
3
3
(a) 3018
(b) 3017.5
(c) 3017.5 –
1 Ê 1 ˆ 2011
1 Ê 1 ˆ 2011
(d)
3018
–
Á ˜
Á ˜
4 Ë 3¯
4 Ë 3¯
Ans. (c)
Solution: Note that
S = 2012 +
1
1
1
1
(2011) + 2 (2010) + 3 (2009) + ... + 2011
3
3
3
3
This is an A.G.P. with a = 2012, b = 1, d = – 1, r = 1/3, n
= 2012
Thus,
S=
–
2012 (-1)(1)(1 - (1 /3)2011 )
+
1 - 1 /3
1 - 1 /3
Solution: We have a4 + a2b2 + b4= (a2 + b2)2 – a2b2
= (a2 + ab + b2) (a2 – ab + b2) etc.
Thus, dividing (2) by (1), we get
(a2 – ab + b2) (b2 – bc + c2) (c2 – ca + a2) = a2b2c2
(3)
But
A.M. ≥ G.M.
1 2
fi
(x + y2) ≥ x2y2
2
and the equality holds if and only if x = y.
fi
x2 – xy + y2 ≥ xy " x, y > 0
and the equality holds if and only if x = y.
Thus, equality in (3) can hold if and only if a = b = c.
Putting this in (1), we obtain
(3a2)3 = a3 fi a3 = (1/3)3 or a = 1/3.
\
a = b = c = 1/3.
Example 84: Suppose a, b, c are distinct positive real
numbers such that a, 2b, 3c are in A.P. and a, b, c are in G.P.
The common ratio of G.P. is
(a) 2
(b) 1/2
(c) 1/3
(d) 3
Ans. (c)
Solution: We have 2(2b) = a + 3c and b2 = ac
[2012 + (2012 - 1)(- 1)](1)(1 /3)2012
(1 - 1 /3)2
fi
= 3018 –
1 È Ê 1ˆ
Í1 - Á ˜
2 Î Ë 3¯
˘ 1 Ê 1ˆ
˙ - ÁË ˜¯
˚ 4 3
1
(a + 3c)2 = ac
16
\
fi
fi
fi
2011
a2 + 9c2 + 6ac = 16ac
a2 – 10ac + 9c2 = 0
a π c]
(a – c) (a – 9c) = 0 fi a = 9c [
c
1
1
1
= or r2 = fi r =
a
3
9
9
2010
1 Ê 1 ˆ 2011
= 3017.5 – ÁË ˜¯
4 3
Example 83: Suppose a, b, c are positive real numbers
satisfying the system of equations
(1)
(a2 + ab + b2) (b2 + bc + c2) (c2 + ca + a2) = abc
and (a4 + a2b2 + b4) (b4 + b2c2 + c4) (c4 + c2a2 + a4)
(2)
= a 3b 3c 3
then
(a) a = b = c = 1/3
(b) abc = 1
1
(a + b + c) = (abc)1/2
(c)
3
(d) a, b, c are in G.P.
Ans. (a)
8.19
[
a, b, c > 0]
Example 85: Let Sn denote the sum of the cubes of first
n natural numbers and sn denote the sum of the first n natural
n
numbers. Then
S
 sr
equals
r =1 r
(a)
1
n(n + 1)
2
(c)
1
1 2
n(n + 1)(n + 2) (d)
n (n + 1)2
6
4
(b)
1
(n + 1)(n + 2)
2
Ans. (c)
Solution: We have
1
Sr = 13 + 23 + ... r3 = r2(r + 1)2
4
sr = 1 + 2 + ... + r =
fi
1
r(r + 1)
2
Sr
1
1
= r(r + 1) = (r2 + r)
sr
2
2
8.20
Complete Mathematics—JEE Main
n
S
 sr
fi
fi
fi
fi
=
1 n 2
 (r + r )
2 r =1
=
1 È1
1
˘
n(n + 1)(2 n + 1) + n(n + 1)˙
2 ÍÎ 6
2
˚
=
1
1
n(n + 1)(2 n + 1 + 3) = n(n + 1)(n + 2)
12
6
r =1 r
x(x – 2) = 0 and (y – 2)(y + 1) = 0
x = 2, y = 2
x–y=0
value of S = ÈÎ 1 ˘˚ + ÈÎ 2 ˘˚ + ÈÎ 3 ˘˚ + ... + ÈÎ 2024 ˘˚ is
(a) 59000
(b) 58750
(c) 59730
(d) 65138
Ans. (c)
Solution: For any positive integer k and x, we have
[ x ] = k ¤ k2 £ x < (k + 1)2 ¤ x Œ [k2, k2 + 2k)
\
1 2
6 10
1
S = + 2 + 3 + 4 + ...
3
3
3
3
3
=1+
44
1
100 99 + 99 100
(r + 1) r + r r + 1
r +1 – r
r (r + 1)
99
Thus,
P=
È 1
ÂÍ
r =1Î
Example 88: Let x =
y=
44
k =1
k =1
2
1
(44)(45)(89) + (44)(45)
6
2
= 59730
1
2 1+ 2
is
r
–
=
=
1
r
1
1
r (r + 1)
–
r +1 + r
1
r +1
˘
1
9
=
˙ =1–
r + 1˚
100 10
1
2 2 2 2 ... upto • and
2 + 2 + 2 + 2 + ... upto • , then x – y equals
(a) 0
(c) 2
Ans. (a)
44
=
1
=
for x = 1936, ..., 2024
 (k )(2k + 1) = 2  (k )2 +  (k )
k =1
(a) 1/10
(b) 3/10
(c) 9/10
(d) 1/2
Ans. (c)
Solution: Let tr denote the rth term of the series, then
tr =
for x = 4, 5, 6, 7, 8,
for x = 9, 10, 11, 12, 13, 14, 15,
=
1
4 / 32
4 2
+
= + =2
3 1 - 1/ 3 3 3
Example 87: Sum of the series P =
+ ... +
[ x] = 2
[ x] = 3
S = (1)(3) + (2)(5) + 3(7) + ... + 44(89)
S=3
3 2 +2 3
for x = 1, 2, 3,
Thus,
2
1 4
4
4
S = 1 + + 2 + 3 + 4 + ...
3
3 3
3
3
+
[ x] = 1
[ x ] = 44
Subtracting, we get
1
x, y > 0]
Example 89: If [x] denotes the greatest integer £ x, then
Example 86: The sum to infinity of the series
2 6 10 14
S = 1 + + 2 + 3 + 4 + ... is
3 3
3
3
(a) 4
(b) 6
(c) 2
(d) 3
Ans. (d)
2 6 10 14
Solution: Let S = 1 + + 2 + 3 + 4 + ...
3 3
3
3
fi
[
(b) 1
(d) none of these
Solution: We have
x2 = 2x and y2 = 2 + y
Example 90: A person is to count 4500 currency notes.
Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ... are in an A.P.
with common difference –2, then the time taken by him to
count all the notes is
(a) 125 minutes
(b) 135 minutes
(c) 24 minutes
(d) 34 minutes
Ans. (d)
Solution: Suppose he takes n minutes to count 4500
notes. We have a1 + a2 + ... + a10 = 10(150), a11 = 148 and
a11, a12, ..., an is an A.P. with common difference d = –2.
We are given
a1 + a2 + ... + a10 + a11 + ... + an = 4500
fi
a11 + a12 + ... + an = 3000
n – 10
[a11 + an] = 3000
2
1
(n – 10)[148 + 148 + (n – 11)(–2)] = 3000
fi
2
fi
(n – 10) (148 – n + 11) = 3000
fi
(n –10) (159 – n) = 3000
fi
n2 – 169n + 4590 = 0
fi
n2 – 135n – 34n + 4590 = 0
fi
(n – 135) (n – 34) = 0 fi n = 135, 34
All the notes get counted in 34 minutes.
fi
Progressions
Example 91: A man saves ` 200 in each of the first three
months of his service. In each of the subsequent months his
savings was ` 40 more than the savings of the immediately
previous month. His total savings from the start of service
will be ` 11040 after
(a) 21 months
(b) 18 months
(c) 19 months
(d) 20 months
Ans. (a)
Solution: Suppose his total savings is ` 11040 after n
months. His savings are
200, 200, 200, 240, 280, º upto n terms
Note that
200, 240, 280, º upto (n – 2) terms
is an A.P. whose sum is 10640.
Thus,
1
(n – 2) [2(200) + (n – 3) (40)] = 10640
2
fi (n–2) (n+7) = 532
fi n2 + 5n – 546 = 0 fi n = –26, 21
As n > 0, n = 21.
Example 92: Let an be the nth term of an A.P.
100
If
100
 a2r
= a and
r =1
 a2r - 1
= b, then the common differ-
r =1
ence of the A.P. is:
1
(a – b)
100
1
(a – b)
(d)
200
(a) a – b
(b)
(c) b – a
Ans. (b)
Solution: Let d the common difference of the A.P., then
a–b=
fid=
100
100
r =1
r =1
 (a2r - a2r - 1 ) =  d = 100 d
1
(a - b ) .
100
Example 93: If m times the mth term of an A.P. with
non-zero common difference equals n times the nth term
of the A.P., where m π n, then (m + n)th term of this A.P. is
(a) (m + n) times mth term
(b) zero
(c) m + n
(d) – (m + n)
Ans. (b)
Solution: Let a be the first term and d be the common
difference of the A.P. We thve
mam = nan
fi
m [a + (m – 1)d ] = n [a + (n – 1)d ]
fi
(m – n) (a – d) + (m2 – n2)d = 0
fi
fi
fi
(m – n) [a – d + (m + n)d ] = 0
a + (m + n – 1) d = 0
[
am + n = 0
8.21
m π n]
Example 94: Let a1, a2, º be in H.P. with a1 = 5 and a20
= 25. The least positive integer n for which an < 0 is
(a) 22
(c) 24
Ans. (d)
Solution:
(b) 23
(d) 25
1 1 1
, , , º are in A.P.
a1 a2 a3
Let d be the common difference of this A.P., then
1
1
= 19d
a20 a1
1 1
4
- = 19d fi d = –
25 5
475
fi
1 ( n - 1) 4
<0
5
475
fi
n – 1 > 95/4 fi n ≥ 25.
Thus, least value of n is 25.
an < 0 fi
Now,
Example 95: A species has an initial population 410.
At the end of first day, the population increases by 50%. At
the end of second day, it decreases by the same percentage.
If the process continues in the same pattern, the number of
days for the population to reach 310 is
(a) 10
(b) 20
(c) 50
(d) 100
Ans. (b)
Solution: Let an be the population at the end of nth
day, then
Ï 10 Ê 3 ˆ n / 2 Ê 1 ˆ n / 2
if n is even
Ô 4 ÁË ˜¯
ÁË ˜¯
2
2
Ô
an = Ì
(n + 1) / 2
(n - 1) / 2
Ô 10 Ê 3 ˆ
Ê 1ˆ
4
is n is odd
˜
Á
˜
Á
ÔÓ
Ë 2¯
Ë 2¯
Ï 410 (3 4 )n / 2 if n is even
Ô
(n - 1) / 2
= Ì
Ê 3ˆ
10 Ê 3 ˆ
Ô4 ÁË ˜¯
ÁË ˜¯ if n is odd
4
2
Ó
As all answers are even natural numbers we set an = 310
assuming n to be even.
Ê 3ˆ
410 Á ˜
Ë 4¯
fi
n2
= 310 fi n = 20
Remark
an = 310 is not possible for any odd n.
8.22
Complete Mathematics—JEE Main
Example 96: The sum of first three terms of an increasing A.P. is 9 and sum of their squares is 35. The sum to n
terms of the series is
(b) 2n2
(a) 3n2
(c) 6n – 2n2
(d) n2
Ans.(d)
Solution: Note that the nth group contain (2n – 1) natural number and the last term of the nth group is
1 + 3 + 5 + º + (2n – 1) = n2
\ First terms of nth group is (n – 1)2 + 1
Thus, sum of first and last term of nth group is
n2 + (n – 1)2 + 1 = 2(n2 – n + 1)
Solution: Let d be common difference of the A.P., then
(a2 – d) + a2 + (a2 + d) = 9
and
(a2 – d)2 + a22 + (a2 + d)2 = 35
fi
a2 = 3 and 3a22 + 2d2 = 35
fi
d = 2.
n
n
Sn = [2a1 + (n – 1)d ] = [2a2 + (n – 3)d ]
2
2
n
[2(3) + (n – 3) (2)] = n2
=
2
Example 98: In an arithmetic progression of 16 distinct
terms with a1 = 16, the sum is equal to square of the last
term. The common difference of the A.P. is
(a) 8/15
(b) – 4/5
(c) – 8/5
(d) – 8/15
Ans. (c)
Example 97: The series of natural numbers is divided
into groups (1), (2, 3, 4), (5, 6, 7, 8, 9), and so on. The sum
of first and last term of the nth group is
(b) n2 – 2n + 2
(a) 2 (n2 – n + 1)
2
(d) n2 – n + 2
(c) n + n
Ans. (a)
Solution: We are given
16
2
[a1 + a16] = a16
2
2
fi
a16
– 8a16 – (8) (16) = 0
fi
(a16 – 16) (a16 + 8) = 0
As
a16 π 16 = a1, a16 = – 8
Now, a16 – a1 = – 24 fi 15d = – 24
fi
d = – 8/5
Assertion-Reason Type Questions
a–x a–y a–z
=
=
.
rz
px
qy
Statement-1: If p, q, r are in A.P., then x, y, z are in H.P.
Statement-2: If p, q, r are in H.P., then x, y, z are in G.P.
Ans. (c)
Example 99: Suppose
Solution: Suppose
a–x
a–y a–z
=
=
= k (say)
px
rz
qy
a–x
a–y
a–z
,q=
,r=
kx
ky
kz
Now, p, q, r are in A.P.
fi
p=
fi
a–x a–y a–z
,
,
are in A.P.
kz
kx
ky
fi
a
a
a
– 1, – 1, – 1 are in A.P.
x
y
z
fi
a a a
, , are in A.P.
x y z
fi
x, y, z are in H.P.
Next, p, q, r are in H.P.
kx
ky
kz
,
,
fi
are in A.P.
a–x a–y a–z
fi
fi
x
y
z
,
,
are in A.P.
a–x a–y a–z
x
y
z
+ 1,
+ 1,
+ 1 are in A.P.
a–x
a–y
a–z
fi a – x, a – y, a – z are in H.P.
Nothing can be said about x, y, z.
Example 100: Let a, b, c, d, e be five non-zero numbers
such that a, b, c are in A.P., b, c, d are in G.P, and c, d, e
are in H.P.
Statement-1: If a = 2, e = 18, b > 0, then b = 4 and c = 6.
Statement-2: If a = 2, e = 18, c = – 6, then b < 0.
Ans. (b)
2ce
Solution: a + c = 2b, c 2 = bd and d =
.
c+e
2+c
36c
,d=
.
Now,
a = 2, e = 18 fi b =
2
c + 18
1
Ê 36c ˆ
\
c 2 = (2 + c ) Á
Ë c + 18 ˜¯
2
fi
fi
c(c + 18) = 18(c + 2)
c 2 = 36 fi c = ± 6.
Progressions
2+6
=4
2
2–6
If c = – 6, then b =
= – 2 < 0.
2
Example 101: Suppose a, b, c, d are four real numbers and
x+a x+b x+a–c
x –1 ,
D(x) = x + b x + c
x+c x+d x–b+d
If c = 6, then b =
Statement-1: If a, b, c, d are in A.P. and
2
Ú0
D(x)dx = – 4,
then common difference of the A.P. is ± 1
Statement-2: If a, b, c, d are in A.P., then D(x) is independent of x.
Ans. (b)
Solution: Let D be the common difference of the A.P.
Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get
x+a x+b x+a–c
D
D
D(x) =
2D – 1
D
D
2D + 1
Again using C1 Æ C1 – C2, we get
x+b x+a–d
D
2D – 1
D
2D + 1
–D
D(x) = 0
0
Now,
2
Ú0
(– 2D2)dx = – 4
fi
4D2 = 4
= – 2D2
fi
2
1
2 3 4
5 6 7 8 9
ººººººº
Statement-1: Sum of the numbers in the 10th row is a number which can be written as sum of two cubes
in two different ways.
Statement-2: Sum of the numbers in the r th row is
1 3
(r + (r + 1)3) if r ≥ 2.
3
Ans. (c)
Solution: The rth row contains (2r – 1) numbers and
its last term is r2.
When read from right to left the rth row forms an A.P. with
first term r2 and common difference – 1.
\ sum of the terms in the rth row
2r – 1
=
[2(r 2) + (2r – 1 – 1) (– 1)]
2
= (2r – 1) (r 2 – r + 1) = r 3 + (r – 1)3
Sum of the terms in the 10th row
= 103 + 93 = 1729 = 123 + 13
Example 104: Statement-1: The sum of the series
1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16)
+ º + (361 + 380 + 400) is 8000
n
2
2
Example 102: Suppose a , b , c are in A.P.
a
b
c
,
,
are in H.P.
Statement-1:
b+c c+a a+b
Statement-2: b + c, c + a, a + b are in H.P.
Ans. (d)
2
Example 103: The natural numbers are divided into
rows as follows:
D=±1
Also, D(x) is independent of x.
2
8.23
Statement-2:
 (k 3 - (k - 1)3 ) = n3
k =1
for each natural number n.
Ans. (a)
Solution: Statement-2 is true since
n
 (k 2 - (k - 1)3 ) = (13 – 03) + (23 – 13) + …
2
Solution: a , b , c are in A.P.
fi
a2 + bc + ca + ab, b2 + bc + ca + ab, c2 + ab + bc
+ ca are in A.P.
fi
(a + b) (a + c), (b + a) (b + c), (c + a) (c + b) are
in A.P.
Dividing by (b + c) (c + a) (a + b), we get
1
1
1
,
,
are in A.P.
(1)
b+c c+a a+b
fi
fi
b + c, c + a, a + b are in H.P.
Putting n = 20, we get
and
a+b+c a+b+c a+b+c
,
,
are in A.P.
a+b
b+c
c+a
fi
a
b
c
+ 1,
+ 1,
+ 1 are in A.P.
b+c
c+a
a+b
fi
a
b
c
,
,
are in A.P.
b+c c+a a+b
k =1
+ (n3 – (n – 1)3) = n3
n
 (k - (k - 1)) ((k 2 + k (k - 1) + (k - 1)2 ) = n3
k =1
n
fi
 (k 2 + k (k - 1) + (k - 1)2 ) = n3
k =1
20
 ((k - 1)2 + k (k - 1) + k 2 ) = 203
k =1
fi
1 + (1 + 2 + 4) + (4 + 6 + 9) + … +
(361 + 380 + 400) = 8000
Thus, Statement-1 is also true and Statement-2 is a correct
reason for it.
8.24
Complete Mathematics—JEE Main
Example 105: Statement-1: For every natural number
n ≥ 2,
1
1
+
1
2
+... +
1
n
Also, for n ≥ 2
1
1
> n
1
>
Statement-2: For every natural number n ≥ 2,
n(n + 1) < n + 1
Ans. (b)
1
+
n
+... +
2
+
1
1
n
+ ... +
n
1
n
n times
1
1
Solution: As n < n + 1 " n ≥ 2
fi
n(n + 1) < (n + 1)2 " n ≥ 2
n(n + 1) < n + 1 " n ≥ 2
fi
Thus, Statement-2 is true.
+
+... +
1
> n
n
1
2
\ Statement-1 is true.
Thus, both the statements are correct but Statement-2 is not
a correct explanation of Statement-1.
fi
LEVEL 2
Straight Objective Type Questions
Example 106: If (1– y) (1 + 2x + 4x2 + 8x3 + 16x4 +
32x5)= 1 – y6, (y π 1), then a value y/x is
(a) 1/2
(b) 2
(c) 1/4
(d) 4
Ans. (b)
Solution: We can rewrite the given expression as
È1 - (2 x )6 ˘˚
= 1 – y 6,
(1 – y) Î
1 - 2x
one of the possible values of y is clearly 2x. Therefore, one
of the possible values of y/x is 2.
22
Example 107: If
and p appear as two distinct terms
7
of an A.P., then common difference of the A.P. must be
(a) an integer
(b) a rational number
(c) an irrational number
(d) 0
Ans. (c)
Solution: Let the A.P. be a1, a2, ... suppose
22
am =
and an = p
7
If d is the common difference of the A.P., then
22
(m – n)d = am – an =
–p
7
1
Ê 22
ˆ
fi d=
– p¯
m–n Ë 7
which is a irrational number as p is an irrational number.
Example 108: Let am = 111 ... 1 , then which of the following is a prime number?
(a) a4
(c) a91
Ans. (b)
m times
(b) a5
(d) a100
Solution: Show that a5 is not divisible by any prime
upto 105.
p /4
Example 109: Let In =
I5, I4 + I6, I5 + I7, …are in
(a) A.P.
(c) H.P.
Ú
tan n x dx. Then I2 + I4, I3 +
0
(b) G.P.
(d) none of these
Ans. (c)
Solution: We have for r ≥ 2,
p /4
tanr x (1 + tan 2 x ) dx
Ir + Ir + 2 =
Ú0
=
Ú0
=
p /4
1
1
˘
tanr +1 x ˙ =
r +1
r
+1
˚0
p /4
tanr x sec 2 x dx
Thus, the given sequence becomes,
1 1 1 1
, , , ,º
3 4 5 6
This is clearly an H.P.
Example 110: If a, b, c are in A.P. p, q, r are in H.P. and
p r
ap, bq, cr are in G.P., then + is equal to
r p
Progressions
(a)
a c
+
c a
(b)
a c
c a
(c)
b q
+
q b
(d)
b a
q p
Adding above expressions, we get
t1 – ntn = 3(t2 + t3 + º + tn)
fi
4t1 – ntn = 3(t1 + t2 + º + tn)
1
1
= 3Sn
1 ◊ 2 ◊ 3 ( n + 1) ( n + 2 ) ( n + 3)
fi
Ans. (a)
Solution: According to the given condition
2 pr
2b = a + c, q =
and b2q2 = (ap) (cr)
p+r
Substituting values of b and q from first two expressions in
the last expression, we get
2
2
(a + c) Ê 2 pr ˆ
ÁË p + r ˜¯ = (ac) (pr)
4
(a + c)
ac
fi
2
( p + r)
pr
=
2
fi
(a)
(c)
1 1
24 n
1
2
n
-
1
n
2
(b)
1
1
18 3 ( n + 1) ( n + 2 ) ( n + 3)
(d)
1
1
18 3 ( n + 3) ( n + 4 )
Solution: Let tr denote the rth term of the given series,
then
1
tr =
and
r (r + 1) (r + 2 ) (r + 3)
tr + 1 =
rtr =
(r + 4)tr + 1 =
Thus,
fi
Putting
Sn =
1 È1
1
˘
Í
3 Î 6 ( n + 1) ( n + 2 ) ( n + 3) ˙˚
Example 112: Sum to n terms of the series
+
1 1! 2!
+
+
5! 6! 7!
3!
+ º is
8!
n! ˆ
1Ê 1
Á ˜
4 Ë 4! (n + 4)!¯
(a)
2
1
5! (n + 1)!
(b)
(c)
1Ê 1
3! ˆ
ÁË ˜
4 3! (n + 2)!¯
(d) none of these
Ans. (b)
Solution: We have tr =
Now, rtr – (r +5)tr+1 =
fi
(r - 1)!
r!
and tr +1 =
(r + 4)!
(r + 5)!
r!
r!
=0
(r + 4)! (r + 4)!
rtr – (r + 1)tr+1 = 4tr+1
n -1
n -1
r =1
r =1
fi 4 Â tr +1 = Â[rtr – (r + 1) tr + 1]
Ans. (b)
fi
fi
a c
p r
+ = +
c a
r p
1
+
Example 111: Sum to n terms of the series
1◊ 2 ◊ 3 ◊ 4
1
1
+
+ is
2◊3◊4◊5 3◊4◊5◊6
8.25
fi
4(t2+ t3 +…+ tn) = 1t1 – ntn
fi
4(t1 + t2 + …tn) = 5t1 – ntn
n!
1
Ê 0!ˆ n(n - 1)!
= 5Á ˜ = Ë 5!¯ (n + 4)!
4! (n + 4)!
1
(r + 1) (r + 2) (r + 3) (r + 4)
1
(r + 1) (r + 2) (r + 3)
fi t1+ t2 +…+ tn =
and
1
(r + 1) (r + 2 ) (r + 3)
rtr – (r + 4)tr + 1 = 0
rtr – (r + 1)tr + 1 = 3tr + 1
r = 1, 2, º, n – 2, n – 1, we get
1t1 – 2t2 = 3t2
2t2 – 3t3 = 3t3
3t3 – 4t4 = 3t4
º
(n – 2)tn – 2 – (n – 1)tn – 1 = 3tn – 1
(n – 1)tn – 1 – ntn = 3tn
n! ˘
1È1
Í
4 Î 4! (n + 4)! ˙˚
Example 113: Sum to n terms of the series
13 + 23 + 33
+ º is
1+ 3 + 5
n
(n2 + 9n + 13)
(a)
24
(c)
(b)
13 13 + 23
+
+
1
1+ 3
n
(2n2 + 7n + 15)
24
n
n
(2n2 + 9n + 13) (d)
(n2 + 11n + 11)
24
24
Ans. (c)
Solution: Let tr denote the rth term of the series, then
1 2
r (r + 1)2
13 + 23 + º + r 3
4
=
tr =
1 + 3 + º + (2r - 1)
r2
8.26
Complete Mathematics—JEE Main
=
1
(r + 1)2
4
fi
=
1 n
1 È n +1 2 ˘
2
(
r
+
)
=
1
Â
ÍÂ r - 1˙
4 r =1
4 Î r =1
˚
fi
1 È (n + 1) (n + 2) (2n + 3) ˘
- 1˙
4 ÍÎ
6
˚
fi
n
 tr
fi
r =1
=
1
[2n3 + 9n2 + 13n + 6 – 6]
24
n
(2n2 + 9n + 13)
=
24
=
n
Example 114: Sum of the series
r
 (r + 1)!
1
n!
(b) 1 -
1
(c) 2 (n + 1)!
Ans. (b)
n -1
n -1
r =1
r =1
 [rtr - (r + 1)tr +1 ] + 2 ÊÁË
t1 – ntn + 1 -
fi
n -1
1
1 ˆ
=
2
˜
 tr +1
r + 1 r + 2¯
r =1
2
= 2(t2 + t3 +…+ tn)
n +1
3t1 – ntn +1 –
n
2
= 2 Â tr
n +1
r =1
n
1
n(2n - 1)
n -1
+
= 2 Â tr
2 n(n + 1)(n + 2) n + 1
r =1
is
r =1
(a) 1 -
1 ˆ
Ê 1
rtr – (r + 1)tr +1 + 2 Á
= 2tr +1
Ë r + 1 r + 2 ˜¯
1
(n + 1)!
(d) none of these
n
\ 2 Â tr =
r =1
(n + 1)(n + 2) - 2(2 n - 1) + 2(n - 1)(n + 2)
2 (n + 1)(n + 2)
n
n(3n + 1)
 tr = 4(n + 1)(n + 2)
fi
r =1
Example 116: If
2
48
47
46
+
+
+ ... +
(48)(49)
(2)(3) (3)(4) (4)(5)
Solution: Let tr denote the rth term of the series, then
tr =
n
n
1
1
1
 tr =  ÊÁË r ! - (r + 1)!ˆ˜¯ = 1 - (n + 1)!
r =1
r =1
fi
Example 115: Sum to n terms of the series
1
3
5
7
+
+
+
+ º is
1.2.3 2.3.4 3.4.5 4.5.6
(a)
n(n + 1)
2(n + 2)(n + 3)
(b)
Ans. (b)
Solution: We have tr =
tr+1 =
fi
rtr =
=
49
We have
50 - r
, r ≥ 2.
r (r + 1)
50 - r
 r (r + 1)
r=2
49
1 ˆ 49 1
Ê1
- Â
= 50 Â Á Ë
r + 1˜¯ r = 2 r + 1
r=2 r
49
1
Ê1 1 ˆ
= 50 Ë - ¯ - Â
2 50
r =2r +1
2r + 1
(r + 1)(r + 2)(r + 3)
50
= 24 -
2r - 1
2r + 1
, (r + 3)tr +1 =
(r + 1)(r + 2)
(r + 1)(r + 2)
-2
(r + 1)(r + 2)
(b) – 1
(d) 1
Solution: Let tr =
2r - 1
and
r (r + 1)(r + 2)
rtr – (r + 3)tr +1 =
51
1 1
1ˆ
Ê
+ k Ë1 + + + . . . + ¯
k
2 3
50
then k is equal to
(a) 2
(c) – 1/2
Ans. (b)
n(3n + 1)
4(n + 1)(n + 2)
1
5
(c)
(d) none of these
6 (n + 1)(n + 4)
fi
+
r
r +1-1 1
1
= =
(r + 1)! (r + 1)! r ! (r + 1)!
1
r =1
=
\
51 50 1
-Â
2 r =1r
k=–1
1
 r + ÊË1 + 2 ˆ¯
1
(49)(50)
Progressions
8.27
EXERCISES
Concept-based
Straight Objective Type Questions
n
È
Ê 7ˆ ˘
1. Let an = Ílog Ë ¯ ˙ and bn = log5 (an) " n Œ N.
Î
5 ˚
Then b1, b2, b3, . . . are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.
2. Suppose a, b, c are positive and form an increasing
G.P. If a + b + c = xb, then
(a) x > 3
(b) x > 4
(d) x > 3 + 5
(c) x > 5
1
3. If for each n ΠN, Sn = nA + n(n + 1) B is sum of
2
the first n terms of an A.P., the common difference of
the A.P. is
(a) A + B
(b) A + 2B
(c) B
(d) B/2
4. An A.P. consists of 2n terms. If two middle terms are
a – b and a + b, then sum of the 2n terms of the
A.P. is
(a) 2n a
(b) 2n b
(c) 2n(a + b )
(d) 2n (a – b )
5. If nth terms of the arithmetic progressions
3, 10, 17, 24, . . .
and 63, 65, 67, 69, . . .
are equal, then n is equal to
(a) 71
(b) 37
(c) 11
(d) 13
6. Suppose a, c ΠQ and b ΠR РQ, then which one
of the following can be true?
(a) a, b + 2, c are in A.P. (b) a, b + 1, c are in G.P.
(c) a, b – 3, c are in H.P. (d) a, b – 2, c are in A.P.
7. Suppose a, b, c > 0, x, y, z > 1 and a, b, c are in
G.P. If alogx = blogy = clogz, then
(a) x, y, z are in A.P.
(b) x, y, z are in G.P.
(c) x, y, z are in H.P.
(d) none of these
8. The ratio of the sum of first three terms of a G.P. to
the sum of first 6 terms of the G.P. is 125 : 152, then
common ratio of the G.P. is
2
3
(b)
(a)
3
4
3
(c)
5
2
(d)
5
9. There are four numbers of which the first three are
in G.P. and the last three are in A.P. with common
difference 9. If the first and last number are equal,
the sum of the four numbers is
(a) 18
(b) 21
(c) 24
(d) 27
10. Suppose A.M. of two positive numbers be 7 and G.M.
between them be 5. The A.M. between their squares
is
(a) 57
(b) 73
(c) 78
(d) 37
11. Sum of an infinite G.P. is 2 and sum of their cubes
is 24, then 5th term of the G.P. is
(a) 3/16
(b) 3/8
(c) – 3/8
(d) – 3/16
12. Suppose a, b are two positive numbers. The product
of three arithmetic means inserted between a, b is
15/2 and product of three arithmetic means inserted
between 1/a and 1/b is 5/18. Then ab is equal to:
(a) 9
(b) 6
(c) 3
(d) 1
1
1
a+b
b+c
, b,
are in A.P., then , b, are in
1 - ab
1 - bc
a
c
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.
13. If
14. The sum of the infinite series
5
55
555
+
+
+...
2
17 (17)
(17)3
is
(a)
1
17
(b)
85
112
(c)
85
324
(d)
19
289
15. Sum of the series
2016
 (-1)r (a + rd )
r =1
is
(a) a + 2016d
(c) 0
(b) 2015d
(d) 1008d
8.28
Complete Mathematics—JEE Main
LEVEL 1
Straight Objective Type Questions
16. If 5a – b, 2a + b, a + 2b are in A.P. and (a – 1)2,
(ab + 1), (b + 1)2 are in G.P., a π 0, then a is equal
to
(a) 2, – 1/4
(b) –2, – 1/4
(c) – 2, 1/4
(d) 2, 1/4
1 1
1
1
+
= 0,
17. If a + c π b and + +
a c a-b c-b
then a, b, c are in
(a) A.P.
(c) H.P.
(b) G.P.
(d) A.G.P.
( p + q) p
(d)
1- p
(c)
a b
(b)
a+ b
ab
(d)
a+ b
2a b
a+ b
1
a+ b
20. Product of two positive integers a and b is 192. Let
g = hcf (a, b) and l = lcm (a, b). If the ratio of A.M.
169
,
between g and l to the H.M. between g and l is
48
then smaller of a, b is
(a) 8, 24
(c) 4, 12
(b) 12, 16
(d) 6, 24
21. The least positive integer n such that 1 –
–...–
(a) 4
(c) 6
2
3
n -1
<
+
1
. Then
(2 ) - 1
(a) a (200) > 100
(c) a (50) < 25
1 1 1
+ + +º
2 3 4
n
(b) a (100) > 100
(d) none of these
4
44 444
26. Let S =
+ 2 + 3 + º upto •. Then S is equal
19
19
19
to
2 ( p + q)
1- p
19. If both the A.M. between m and n and G.M. between
two distinct positive numbers a and b are equal to
ma + nb
, then n is equal to
m+n
(a)
24. If cos (x – y), cos x and cos (x + y) are in H.P. then
value of cos x sec (y/2) is
(b) ± 3
(a) ± 2
(c) ± 2
(d) + 1
25. For a positive integer n, let a (n) = 1 +
18. In an A.P., the first term is 1 and sum of the first p
terms is 0, then sum of the first (p + q) terms is
q
q ( p + q)
(b)
(a)
1- p
1- p
(c)
23. The harmonic mean of the roots of the equation
(5 + 2 )x2 – (4 + 2 )x + 8 + 2 2 = 0 is
(a) 2
(b) 4
(c) 6
(d) 8
2 2
3 32
1
is
100
(b) 5
(d) 7
22. Let the positive numbers a, b, c, d be in A.P. Then
abc, abd, acd, bcd are
(a) Not in A.P./G.P./H.P. (b) in A.P.
(c) in G.P.
(d) in H.P.
(a) 38/81
(c) 36/171
(b) 4/19
(d) none of these
27. If a, b, c, …are in G.P., and a1/x = b1/y = c1/z = …,
then x, y, z, … are in
(a) H.P.
(b) G.P.
(c) A.P.
(d) none of these
28. If ax = by = cz = du and a, b, c, d are in G.P., then
x, y, z, u are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) none of these
29. If a, b, c, d and p are distinct real numbers such that
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2)
£ 0, then a, b, c, d are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) none of these
30. If a, b, c are in A.P., then 2ax +1, 2bx +1, 2cx +1, x ΠR,
are in
(a) A.P.
(b) G.P. and when x > 0
(c) G.P. only when x < 0 (d) G.P. for all x
31. The rational number which is equal to the number
2.357 with recurring decimal is
(a)
2355
1001
(b)
2370
999
(c)
2355
999
(d)
2359
991
Progressions
32. The sum to infinity of the series 1+
4 7 10
+
+ +º
5 52 53
40. If l, m, n are positive and are respectively the pth, qth
and rth terms of a G.P., then
is
16
11
(b)
35
8
35
17
(c)
(d)
16
6
33. If the sum to n terms of an A.P. is 3n2 + 5n, while
Tm = 164, then value of m is
(a) 25
(b) 26
(c) 27
(d) 28
(a)
34. If G1 and G2 are two geometric means and A is the
arithmetic mean inserted between two positive numG2 G2
bers then the value of 1 + 2 is
G2 G1
(a) A/2
(c) 2A
(b) A
(d) none of these
35. If A1, A2 be two arithmetic means and G1, G2 be two
geometric means between two positive numbers a and
A + A2
is equal to
b, then 1
G1G2
ab
(a)
a+b
(c)
a+b
2ab
2ab
(b)
a+b
(d)
a+b
ab
36. The first and last terms of an A.P. are a and l
respectively. If s is the sum of all the terms of the A.P.,
then the common difference of the A.P. is
l 2 - a2
(a)
s - (l + a)
(c)
l 2 + a2
2s - (l + a)
l 2 - a2
(b)
2s - (l + a)
(d)
l 2 + a2
s - (l - a)
37. The value of x = 21/4 41/8 81/16 161/32 … is
(a) 2
(b) 3/2
(c) 1
(d) 1/2
38. The sum of an infinite number of terms of a G.P. is
20, and the sum of their squares is 100, then the first
term of the G.P. is
(a) 5
(b) 8/5
(c) 3/5
(d) 8
39. The eighth term of a geometric progression is 128
and common ratio is 2. The product of the first five
terms is
(b) 45
(a) 46
3
(c) 4
(d) 48
8.29
D=
log l
p 1
log m q 1 is equal to
log n r 1
(a) pqr
(c) p + q + r + pqr
(b) p + q + r
(d) 0
41. Let a1, a2, a3 º be terms of an A.P. If
a1 + a2 + ... + a p
p2
=
a1 + a2 + ... + aq
q
2
a6
equals
a21
, p π q then
11
41
(b)
41
11
7
2
(c)
(d)
2
7
42. The sum to 10 terms of the series
(a)
2 + 6 + 18 + 54 + º is
(a) 121
(
6+ 2
(c) 243
(
3 +1
)
)
121
2
(
3 +1
(d) 243
(
3 -1
(b)
)
)
43. If 12 + 22 + …+ n2 = 1015, then value of n is
(a) 15
(b) 14
(c) 13
(d) 16
44. 12 + (12 + 22) + (12 + 22 + 32) + … upto 22nd terms
is
(a) 23276
(b) 22736
(c) 22738
(d) 227152
45. Suppose a, b, c are distinct real numbers. If a, b, c
are in A.P. and a2, b2, c2 are in H.P., then
a
(a) – , b, c are in G.P.
2
(b) a + b = c
(c) a = b + c
(d) a, b, c are in G.P.
46. If mth term of an A.P. is n and nth term in m, then
its rth term is
(a) m + n – r
(b) m + n + r
(c) m – n + r
(d) r – (m + n)
47. If a, b, c are positive real numbers that are in G.P.,
then the equations ax2 + 2bx + c = 0 and dx2 + 2ex
+ f = 0 have a common root if a/d, b/e, c/f are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) none of these
8.30
Complete Mathematics—JEE Main
48. If log102, log10(2x + 1) and log10(2x + 3) are in A.P.,
then
(a) x = 0
(b) x = 1
1
1
log102
(d) x = log25
2
2
1
49. Suppose x , |x + 1|, |x – 1| are in A.P., then sum to
2
10 terms of the A.P. is
(a) 54
(b) 36
(c) 28
(d) none of these
(c) x =
50. The harmonic mean and geometric mean of two positive numbers are in the ratio 4 : 5, then two numbers
are in the ratio
(a) 4 : 1
(b) 3 : 1
(c) 2 : 1
(d) 1 : 3
51. The sum to 16 terms of the series
3
13 +
3
3
3
3
3
3
3
1+ 2 1 + 2 + 3 1 + 2 + 3 + 4
+ º is
+
+
1+ 3
1+ 3 + 5
1+ 3 + 5 + 7
(a) 445
(c) 447
(b) 446
(d) 448
52. If log (a + c) + log(a + c –2b) = 2 log (a – c), then
(a) a, b, c are in A.P.
(b) a, b, c are in G.P.
(c) a, b, c are in H.P.
(d) b + c, c + a, a + b are in H.P.
53. The sum of integers from 1 to 100 that are divisible
by 2 or 5 is
(a) 3050
(b) 3150
(c) 3250
(d) 3350
54. The common difference d of the A.P. in which T7 = 9
and T1T2T7 is least is
(a) 33/2
(b) 5/4
(c) 33/20
(d) 7/3
55. If 3 2sin 2q – 1, 14, 34–2sin 2q are first three terms of an
A.P., then its fifth term is given by
(a) 25
(b) 40
(c) 53
(d) – 12
3 5 9 17
+ + + +º
2 4 8 16
56. The sum to n terms of the series
is equal to
(a) n + 1 – 2–n
(c) 2n – 1
(b) 1 + 2–n
(d) 2n – n + 1
2
2
2
2
Ê n ˆ
(a) tan–1 Á
Ë n + 2 ˜¯
Ê 2 n - 1ˆ
(b) tan–1 Á
Ë 2n + 1˜¯
Ê 1ˆ
(c) tan–1 Á ˜
Ë 3n ¯
Ê 1ˆ
(d) tan–1 Ë ¯
2n
59. Sum to 30 terms of the series
1
1
1
+
+
+º
1.2.3 2.3.4 3.4.5
is
(a)
116
465
(b)
495
1984
(c)
435
791
(d)
485
791
60. If n!, 3(n!) and (n + 1)! are in G.P., then n!, 5(n!)
and (n + 1)! are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) none of these
61. If Sn = 81 + 54 + 36 + 24 +… upto n terms, then
value of
x=
Sn - 4Sn -1 + 6Sn - 2 - 4Sn -3 + Sn - 4
is equal
Sn -1 - 4Sn - 2 + 6Sn -3 - 4Sn - 4 + Sn - 5
2
3
(b)
3
2
1
(c)
(d) 2
2
62. The sum to n terms of the series 12 + (1) (2) + 32 +
(3) (4) + 52 + (5) (6) + 72 + … upto n terms when
n is odd is
1
(a)
(n + 1) (4n2 – n + 3)
12
1
(b)
n(4n2 + 3n – 4)
12
1
(c)
(n + 1) (4n2 – n + 5)
6
(d) none of these
(a)
63. If L = lim (1 + 3–1) (1 + 3–2) (1 + 3–4) (1 + 3–8) …
nƕ
57. The sum of first n terms of the series
2
Ê 1ˆ
58. Sum to n terms of the series tan–1 Á ˜ +
Ë 3¯
1
1
Ê
ˆ
Ê
ˆ
tan–1 Á ˜ + tan–1 Á ˜ +… is
Ë 7¯
Ë 13 ¯
n
2
1 + 2(2 ) + 3 + 2(4 ) + 5 + 2(6 ) + … is
n(n + 1)2/2 when n is even. When n is odd, the sum is
(b) n2(n + 1)/2
(a) n(n – 1)2/2
(c) n(n + 2)/3
(d) n(n + 3)/2
(1 + 3-2 ), then
2
3
(c) L = 1
(a) L =
3
2
(d) L = 2
(b) L =
Progressions
64. Suppose a1, a2,…, an are n distinct odd natural numbers not divisible by any prime greater than 5, then
1
1
1
+
+º
cannot exceed
value of the expression
a1 a2
an
(a) 2
(c) 1/2
(b) 1
(d) none of these
65. Coefficient of x99 in the expansion of
(x – 1) (x – 3) (x – 5)…(x – 199) is
(a) – 100
(b) – 1000
(c) –10000
(d) – 100000
66. Sum to n terms of the series 1 + (1 + 2) + (1 + 2 +
3) +… is
1
1
(b) n(n + 1) (n + 2)
(a) n(n + 1) (n + 2)
2
3
(c)
1
n(n + 1) (2n + 1)
4
•
67. If x =
•
z=
(d)
1
n (n + 1) (n + 2)
6
•
 an , y =
 bn ,
n= 0
n= 0
0 < a < b < 1, and
n
a
 ÊÁË b ˆ˜¯ , then
n=0
(a) x + yz = x(y + z)
(c) xy + z = y(x + z)
(b) xyz = x + y + z
(d) none of these
68. If the sum of an infinite decreasing G.P. is 3 and sum
of the cubes of its terms is 108/13, then common ratio
is given by
1
1
(a)
(b)
5
4
1
1
(c)
(d)
3
2
Ê ar +1 ˆ
69. The sum of the series S = Â log Á r -1 ˜ is
Ëb ¯
n
r =1
(a)
Ê a n -1 ˆ
n
log Á n ˜
2
Ë b ¯
(b)
Ê a n +3 ˆ
n
log Á n -1 ˜
2
Ëb ¯
(c)
Ê a n+2 ˆ
n
log Á n -1 ˜
2
Ëb ¯
(d) none of these
70. Fifth term of a G.P. is 2, then the product of its 9
terms is
(a) 256
(b) 512
(c) 1024
(d) 204
71. If s is the sum to infinity of a G.P., whose first term
is a, then the sum of the first n terms of the G.P. is
È Ê
aˆ n ˘
(a) s Í1 - Á 1 - ˜ ˙
s¯ ˚
Î Ë
aˆ
Ê
(b) s Á 1 - ˜
Ë
s¯
È Ê
aˆ n ˘
(c) a Í1 - Á 1 - ˜ ˙
s¯ ˚
Î Ë
(d)
8.31
n
aÈ Ê
aˆ n ˘
1
1
Í Á
˜ ˙
sÎ Ë
s¯ ˚
72. For a sequence <an>, a1 = 5, and
ar +1 1
= " r, then
ar
2
•
 a2 n-1
is
n =1
(a) 20/7
(c) 20/3
(b) 20/5
(d) 20
73. If H.M. and G.M. of two positive numbers are in the
ratio 12 : 13, then the numbers are in the ratio
(a) 1 : 2
(b) 2 : 3
(c) 3 : 5
(d) 4 : 9
74. The H.M. between two numbers is 16/5, their A.M.
is A and G.M. is G. If 2A + G2 = 26 then the numbers are
(a) 6, 8
(b) 4, 8
(c) 2, 8
(d) 1, 8
75. If ar = a + (r – 1)d is rth term of an A.P., then
an2 - 2an2-1 + an2- 2 is independent of
(a) n
(c) a
(b) d
(d) a and n both
76. If A = 1 + r a + r 2a + r 3a + … upto • and B = 1 +
a
r b + r 2b + r 3b +…upto •, then
is equal to
b
(a)
log (1 + A)
log (1 + B)
(c) logB A
(b)
log ( A - 1)
log ( B - 1)
(d) none of these
77. Suppose a1 = 45, a2 = 41 and ak = 2ak – 1 – ak – 2 "
k ≥ 3, then, value of
S=
1 2
2
2
Èa1 – a22 + a32 – a42 + ... + a11
˘˚ is
– a12
12 Î
(a) 89
(c) 91
(b) 90
(d) 92
78. Suppose P(x) = 1 – x + x2 – x3 + ... + x2012 is expressed as a polynomial in y, as Q(y) = a0 + a1y +
2012
... + a2012 y2012 where y = x – 2, then
 ai
i=0
1 2013
(3
+ 1)
4
(c) 0
(a)
1 2013
(3
– 1)
3
(d) 1
(b)
equals
8.32
Complete Mathematics—JEE Main
79. The sum of first 26 terms of
if a2 + a6 + a9 + a18 + a21 +
(a) 705
(b)
(c) 725
(d)
an A.P. a1, a2, a3, ... .
a25 = 165, is
715
735
80. If the sum of first n terms of an A.P. is cn2, then the
sum of squares of these n terms is
(a)
(c)
(
)
n 4n2 – 1 c 2
6
(
)
n 4n2 – 1 c 2
3
(b)
(d)
(
)
n 4n2 + 1 c 2
3
(
)
n 4n2 + 1 c 2
6
Assertion-Reason Type Questions
n
81. Statement-1 : If Sn =
Â
ak = 3n2 + 2n – 7 for each
k =1
n, then, a1, a2, a3, º are in A.P.
Statement-2 : Sum to n terms of an A.P. is always
of the form an2 + bn.
82. Statement-1 : The largest interval for which we can
find the sum of the series
1 + (2x – 1) + (2x – 1)2 + (2x – 1)3 + º is (0, 1).
a + ar + ar2 + º
can be summed up if | r | < 1.
83. Statement-1 : If a, b, c > 0 and the expression
( ( a + c ) + 4b ) x + b > 0
2
Statement-2 : ax2 + bx + c > 0 " x ΠR if and only
if a > 0 and b2 – 4ac < 0.
84. Let a, b be roots of ax2 + 2bx + c = 0 and D =
b2 – ac.
Statement-1 : If a + b, a2 + b 2, a3 + b 3 are in G.P.
then cD = 0
Statement-2 : If a, ab, b are in A.P. then b + c = 0
85. Suppose four distinct positive numbers a1, a2, a3, a4
are in G.P.
Statement-2 : An infinite geometric series
(a + c) x2 +
" x ΠR, then a, b, c are in G.P.
2
Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 and b4 = b3
+ a 4.
Statement-1: The numbers b1, b2, b3, b4 are neither
in A.P. nor in G.P.
Statement-2: The numbers b1, b2, b3, b4 are in H.P.
LEVEL 2
Straight Objective Type Questions
1 7 1 20
86. The nth term of the sequence 2 , 1 , 1 , º is
2 13 9 23
(a)
15
5n + 3
(b)
20
5n + 4
(c)
20
5n + 3
(d)
20
5n + 2
87. H.M. between
1
1
and
is
28
10
(a)
1
19
(b)
1
18
(c)
1
17
(d)
1
16
Ê 1 1ˆ Ê 1 1 ˆ Ê 1 1 ˆ
88. If a Á + ˜ , b Á + ˜ , c Á + ˜ are in A.P., then
Ë b c¯ Ë c a¯ Ë a b¯
(a) a, b, c are in A.P.
(b) a, b, c are in G.P.
(c) a, b, c are in H.P.
(d) none of these
89. If a1, a2, …, an are in A.P., with common difference
d π 0, then sum of the series sin d [cosec a1 cosec
a2 + cosec a2 cosec a3 +…+ cosec an–1 cosec an] is
(b) cosec a1 – cosec an
(a) sec a1 – sec an
(c) cot a1 – cot an
(d) tan a1 – tan an
1 1
1
+ + º + , then value of
n
2 3
2n - 1
3 5
is
1+ + +º+
2 3
n
90. If Hn = 1 +
Progressions
(a) Hn + n
(c) n – 1 + Hn
(b) 2n – Hn
(d) Hn + 2n
a
b
c
91. If a, b, c are in H.P., then
,
,
b+c–a c+a–b a+b–c
are in
(a) A.P.
(c) H.P.
(b) G.P.
(d) A.G.P.
92. Sum of all the terms in the nth row of the triangle
1
2
3
4
5
6
7
8
9
10
.
.
.
.
.
.
.
.
.
.
.
.
.
.
is
1
1
(a) n2(n + 1)
(b) n(n2 + 1)
2
2
(c) n(n2 + 1)
(d) n2(n + 1)
93. Let a and b be two positive real numbers. Suppose A1,
A2 are two arithmetic means; G1, G2 are two geometric
means and H1, H2 are two harmonic means between
a and b, then
A1 + A2
is equal to
H1 + H 2
2(a 2 + b2 ) + 5ab
(a)
9 ab
(c)
(d)
1
1
1
+
+ ... + (– 1)n – 1 , then
2
3
n
1
1
1
+
+ ... +
is equal to
n +1 n + 2
2n
(a) a (2n)
(b) a (2n) – a (n)
(c) a (3n) – a (2n)
(d) none of these
94. Let a (n) = 1 –
95. In Fig. 8.2 each square is filled with the arithmetic
mean of the numbers in the boxes sharing a side with
it. If the first square is filled with 5, then value of x is
(a) 5
(b) 0
(c) – 5
(d) cannot be determined.
5
x
Fig. 8.2
96. Let a, b and c be three real
È1 9
[ a b c] Í8 2
Í
ÍÎ7 3
numbers satisfying
5˘
5˙ = [ 0 0 0 ]
˙
5˙˚
(1)
and let b = 6. If a and b are the roots of the quadratic
a 2 + b2
(b)
+5
9 ab
a 2 + b2 + 5(a + b)
9 ab
8.33
equation ax2 + bx + c = 0, then
a 2 + b2 + 7(a + b)
3 (a + b) ab
(a) 6
(c) 6/7
•
Ê 1 1ˆ
 ÁË a + b ˜¯
n=0
n
is
(b) 7
(d) •
Previous Years' AIEEE/JEE Main Questions
1. Fifth term of a G.P. is 2, then the product of its 9
terms is
(a) 256
(b) 512
(c) 1024
(d) 2048
[2002]
1
1
1
2. The value of 2 4 . 4 8 . 8 16 ... upto • is
(a) 1
(b) 2
3
(c)
(d) 4
2
3. If 1, log9(31 –
x equals
(a) log34
(c) 1 – log43
x
[2002]
+ 2), log3(4(3x) – 1) are in A.P., then
(b) 1 – log34
(d) log43
[2002]
4. 13 – 23 + 33 – 43 + ... + 93 equals
(a) 425
(b) – 425
(c) 475
(d) – 475
[2002]
5. The sum of integers from 1 to 100 that are divisible
by 2 or 5 is
(a) 3000
(b) 3050
(c) 3600
(d) 3250
[2002]
6. Sum of an infinite G.P. is 20 and sum of their squares
is 100. The common ratio of the G.P. is
1
2
(b)
(a)
5
5
3
4
(c)
(d)
[2002]
5
5
8.34
Complete Mathematics—JEE Main
7. Let f(x) be a polynomial function of second degree. If
f (1) = f (– 1) and a, b, c are in A.P., then f ¢(a), f ¢(b),
and f ¢(c) are in
(a) G.P.
(b) H.P.
(c) A.G.P.
(d) A.P.
[2003]
8. Let Tr be the r th term of an A.P. whose first term is
a and common difference is d. If for some positive
1
1
and Tn =
, then a
integers m, n, m π n, Tm =
n
m
– d equals
(a)
1
mn
1 1
+
m n
9. The sum of the first n terms of the series
(d)
[2004]
1
n(n + 1)2
2
when n is even. When n is odd the sum is
1
1
(a) n(n + 1)2
(b) n2(n + 1)
4
2
12 + 2.22 + 32 + 2.42 + 52 + 2.62 + ... is
3
(c) n(n + 1)
2
1
(d) n2(n + 1)2
4
[2004]
10. If a, b, c are in A.P. are | a | < 1, | b | < 1, | c | < 1,
and x =
•
•
•
n=0
n=0
n=0
 a n , y =  bn , z =  cn , then x, y,
z are in
(a) A.P.
(c) H.P.
(b) G.P.
(d) A.G.P.
[2005]
11. If the expression in powers of x of the function
1
is a0 + a1x + a2x2 + a3x3 + ..., then
(1 - ax ) (1 - bx )
an is
(a)
bn + 1 – a n + 1
b–a
(b)
bn – a n
b–a
(c)
a n – bn
b–a
(d)
a n + 1 – bn + 1
b–a
12. Let a1, a2, a3, ... be terms of an A.P. if
a1 + a2 + ... + a p
p2
= 2 , p π q,
a1 + a2 + ... + aq
q
11
41
(d)
2
7
[2006]
14. In a geometric progression consisting of positive
terms, each term equals the sum of the next two terms.
Then common ratio of the G.P. is
1
(1 - 5 )
2
(b)
1
5
2
1
[2007]
( 5 + 1)
2
15. Statement-1: For every natural number n ≥ 2,
1
1
1
+
+
> n
1
2
n
(c)
(d)
5
Statement-2: For every natural number n ≥ 2,
n ( n + 1) < n + 1.
[2008]
16. The sum to infinity of the series S = 1 +
10 14
+
+ ... , is
33 34
(a) 4
(c) 2
2 6
+
3 32
+
(b) 6
(d) 3
[2009]
17. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute.
If a1 = a2 = ... a10 = 150 and a10, a11, ... are in A.P.
with common difference – 2, then the time taken by
him to count all the notes is
(a) 125 minutes
(b) 135 minutes
(c) 24 minutes
(d) 34 minutes
[2010]
18. A man saves ` 200 in each of the first three months
of his service. In each of the subsequent months his
saving is ` 40 more than the saving of the immediately
previous month. His total saving from the start of the
service will be ` 11040 after.
(a) 21 months
(b) 18 months
(c) 19 months
(d) 20 months
[2011]
19. Let an be the nth term of an A.P.
100
If
 a 2r
r =1
100
= a and
 a 2r -1
= b, then
r =1
common difference of the A.P. is
1
(a – b)
(a) a - b
(b)
100
a
then 6 equals
a21
(a)
[2006]
7
2
13. If a1, a2, ... an are in H.P., then the expression a1a2
+ a2a3 + ... + an – 1 an is equal to
(b) n(a1 – an)
(a) (n – 1) a1 an
(d) n a1 an
[2006]
(c) (n – 1) (a1 – an)
(a)
(b) 1
(c) 0
(c)
(b)
41
11
(c) b - a
(d)
1
(a – b)
200
[2011]
Progressions
20. If 100 times the 100th term of an A.P. with non zero
common difference equals the 50 times its 50th term,
then the 150 term of this A.P. is
(a) 150 times its 50th term
(b) 150
(c) zero
(d) –150
[2012]
21. Statement-1 : The sum of the series 1 + (1+2 +4)
+ (4 + 6 + 9) + (9 + 12 + 16) + ... + (361 + 380 +
400) is 800.
n
Statement-2 :
 (k
k =1
3
3
)
- ( k - 1) = n , for any natural
number n
3
[2012]
22. If x, y, z are in A.P. and tan–1 x, tan–1 y and tan–1 z
are also in A.P., then
(a) 2x = 3y = 6z
(b) 6x = 3y = 2z
(c) 6x = 4y = 3z
(d) x = y = z
[2013]
23. The sum of first 20 terms of the sequence 0.7, 0.77,
0.777, . . . is
7
7
(b)
(179 + 10–20)
(a) (99 – 10–20)
9
81
7
(c) (99 + 10–20)
9
7
(d)
(179 – 10–20) [2013]
81
24. Given sum of the first n terms of an A.P. is 2n + 3n2.
Another A.P. is formed with the same first term and
is double the common difference, the sum of n terms
of the new A.P. is
(b) 6n2 – n
(a) n + 4n2
(c) n2 + 4n
(d) 3n + 2n2 [2013, online]
25. The sum
3
5
7
+ 2
+ 2
+ . . . upto 11 terms
2
2
1
1 +2
1 + 22 + 32
is
7
11
(a)
(b)
2
4
11
60
(c)
(d)
[2013, online]
2
11
26. Given a sequence of 4 numbers, first three of which
are in G.P., and the last three are in A.P. with common
difference six. If first and last term of the sequence
are equal, then the last term is
(a) 16
(b) 8
(c) 4
(d) 2
[2013, online]
27. The value of
12 + 32 + 52 + . . . + 252
is
(a) 2925
(b) 1469
(c) 1728
(d) 1456
[2013, online]
8.35
28. Let a1, a2, a3, . . . be an A.P. such that
a1 + a2 + . . . + a p
a1 + a2 + . . . + aq
Then
(a)
=
p2
q2
;pπq
a6
is equal to
a21
41
11
11
41
29. The sum of the series
(c)
(b)
121
1681
(d)
121
1861
[2013, online]
22 + 2(42) + 3(62) + . . . upto 10 terms is
(a) 11300
(b) 11200
(c) 12100
(d) 12300 [2013, online]
1
1
Ê
ˆ
Ê
ˆ
30. If S = tan–1 Á 2
+ tan–1 Á 2
Ë n + 3n + 3 ˜¯
Ë n + n + 1˜¯
1
ˆ
Ê
+ . . . + tan–1 Á
, then
Ë 1 + (n + 19) (n + 20 ) ˜¯
tan (S) is equal to
20
(a)
401 + 20n
(c)
20
2
n + 20 n + 1
(b)
(d)
n
2
n + 20 n + 1
n
401 + 20 n
[2013, online]
31. If a1, a2, a3, . . ., an . . . are in A.P. such that
a4 – a7 + a10 = m, then sum of the first 13 terms of
the A.P. is
(a) 10 m
(b) 12 m
(c) 13 m
(d) 15 m
[2013, online]
32. Three positive numbers form an increasing G.P. If the
middle term in this G.P. is doubled, the new numbers
are in A.P. Then the common ratio of the G.P. is
(a) 3 + 2
(b) 2 - 3
(c) 2 + 3
(d)
2+ 3
[2014]
33. If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 + . . . + 10 (11)9
= k (10)9, then k is equal to
441
(a)
(b) 100
100
221
[2014]
10
34. Given an A.P. whose terms are all positive integers.
The sum of its nine terms is greater than 200 and less
than 220. If the second term in it is 12, then its 4th
term is
(c) 110
(d)
8.36
Complete Mathematics—JEE Main
(a) 8
(c) 20
(b) 16
(d) 24
[2014, online]
35. If the sum
3
5
7
+ 2
+ 2
+ . . . + upto
2
2
1
1 +2
1 + 22 + 32
k
20 terms is equal to
, then k is equal to
21
(a) 120
(b) 180
(c) 240
(d) 60
[2014, online]
36. In a geometric progression, if the ratio of the sum of
first 5 terms to the sum of their reciprocals is 49, and
the sum of the first and the third term is 35. Then the
first term of this geometric progression is
(a) 7
(b) 21
(c) 28
(d) 42
[2014, online]
37. The sum of the first 20 terms common between the
series 3 + 7 + 11 + 15 + . . . and 1 + 6 + 11 + 16
+ . . ., is
(a) 4000
(b) 4020
(c) 4200
(d) 4220
[2014, online]
38. Let G be the geometric mean of two positive numbers
1
1
and .
a and b, and M be the arithmetic mean of
a
b
1
If
: G is 4 : 5, then a : b can be
M
(a) 1 : 4
(b) 1 : 2
(c) 2 : 3
(d) 3 : 4
[2014, online]
2 2
39. The least positive integer n such that 1 – - 2 –
3 3
2
1
. . . – n -1 <
, is
100
3
(a) 4
(b) 5
(c) 6
(d) 7
[2014, online]
40. The number of terms in an A.P. is even: the sum of
the odd terms in it is 24 and that the even terms is
1
30. If the last terms exceeds the first term by 10 ,
2
then the number of terms in the A.P. is
(a) 4
(b) 8
(c) 12
(d) 16
[2014, online]
È 1 3n ˘
41. Let f(n) = Í +
n , where [n] denotes the greatest
Î 3 100 ˙˚
56
integer less than or equal to n. then
 f (n) is equal to
n =1
(a) 56
(c) 1287
(b) 689
(d) 1399
[2014, online]
42. If m is the A.M. of two distinct real number l and
n (l, n > 1) and G1, G2 and G3 are three geometric
between l and n, then G14 + 2G24 + G43 equals,
(b) 4lm2n
(a) 4l2 mn
2
(d) 4l2m2n2
[2015]
(c) 4lmn
43. The sum of first 9 terms of the series
13 13 + 23 13 + 23 + 33
+
+
+
1
1+ 3
1+ 3 + 5
(a) 71
(c) 142
is
(b) 96
(d) 192
[2015]
30
4. The value of  (r + 2)(r - 3) is equal to
r =16
(a) 7785
(c) 7775
(b) 7780
(d) 7770
rd
[2015, online]
th
45. The sum of the 3 and the 4 terms of a G.P. is 60
and the product of its first three terms is 1000. If the
first term of this G.P. is positive, then its 7th term is:
(a) 7290
(b) 320
(c) 640
(d) 2430
[2015, online]
5
k
1
= , then k is equal to
3
n =1 n( n + 1)( n + 2)( n + 3)
46. If Â
55
336
1
(c)
6
17
105
19
(d)
[2015, online]
112
47. If the 2nd, 5th and 9th terms of a non-constant A.P. are
in G.P., then the common ratio of this G.P. is
8
4
(b)
(a)
5
3
(a)
(b)
(c) 1
(d)
2
2
7
4
2
[2016]
2
16
Ê 3ˆ
Ê 2ˆ
Ê 1ˆ
2 Ê 4ˆ
ÁË1 ˜¯ + ÁË 2 ˜¯ + ÁË 3 ˜¯ + 4 + ÁË 4 ˜¯ + ,is m,
5
5
5
5
5
then m is equal to
(a) 102
(c) 100
(b) 101
(d) 99
[2016, online]
49. Let a1, a2, a3, …, an, … be in A.P. If a3 + a7 + a11
+ a15
to
(a) 306
(b) 204
(c) 153
(d) 612
[2016, online]
Progressions
8.37
Previous Years' B-Architecture Entrance
Examination Questions
1
1
, a, b,
are
16
6
in G.P. and the last three are in H.P., then the values
of a and b respectively are
1. If the first three terms of a sequence
(a)
1 4
,
12 9
(b)
4 3
,
7 4
1
(d) - , 1
4
1 1
(c) ,
9 12
[2006]
2. If a, x, b are in H.P. and a, y, z, b are in G.P., then
yz
is
the value of
3
x ( y + z3 )
(a) ab
(b)
1
2ab
1
ab
(d) 2ab
[2006]
2
3. If x = 1 + a + a2 + . . . (|a| < 1) and y = 1 + b + b2
+ . . . (|b| < 1), then sum of the series
(c)
1 + ab + a2b2 + . . . is
xy
(a)
x - y -1
(c)
xy
x + y +1
(b)
xy
x - y +1
(d)
xy
x + y -1
[2007]
4. If three distinct positive numbers a, b, c are in A.P.
such that abc = 4, then value of b is always
(a) greater than (2)2/3
(b) less than (2)2/3
2/3
(c) equal to (2)
(d) equal to (2)3/2 [2007]
5. The value of the sum
20
1
1
1
1
 i ÈÍÎ i + i + 1 + i + 2 + . . . + 20 ˘˙˚
9. A tree, in each year grows 5 cm less than it grew in
the previous year. If it grew half a metre in the first
year, then the height of the tree (in metres) when it
ceases to grow, is
(a) 3.00
(b) 2.75
(c) 2.50
(d) 2.00
[2013]
48
47
46
2
+
+
+ . . . +
+
10. If
(2)(3)
(3)(4)
(4)(5)
(48)(49)
1 1
1ˆ
1
51
Ê
=
+ K Ë 1 + + + . . . + ¯ , then
2 3
50
(49)(50)
2
K equals
(a) –1
13. If  f(k ) =
[2008]
6. The sum of the numbers between 200 and 400 that
are divisible by 7 is
(a) 8729
(b) 7511
(c) 6328
(d) 5712
[2009]
7. Let a, b and c be distinct real numbers. If a, b, c are
in geometric progression and a + b + c = xb, then x
lies in the set
(b) –
1
2
(c) 1
(d) 2
[2014]
11. If log10 2, log10 (2x – 1) and log10 (2x + 3) are three
consecutive terms of an A.P. for
(a) no real x
(b) exactly one real x
(c) exactly two real x
(d) more than two real x
[2014]
12. Let a, b, c, d and and e be real numbers such that
1 1 1
a > b > 0 and c > 0. If , , are in A.P., b, c, d
a b c
ab 2
are in G.P. and c, d, e are in A. P., then
is
( 2a - b) 2
equal to
n
(b) 105
(d) 115
(b) (–1, 0) » (1, 2)
(d) (0, 1)
[2010]
8. If the sum of first n terms of two A.P.’s are in the
ratio 3n + 8 : 7n + 15, then the ratio of 12th term is
(a) 8 : 7
(b) 7 : 16
(c) 74 : 169
(d) 13 : 47
[2012]
(a) c
(c) e
i =1
is
(a) 100
(c) 110
(a) (1, 3)
(c) (– •, –1) » (3, •)
k =1
(b) de
(d) d
[2015]
10 1
2n
, then Â
is equal to
n +1
k =1 f( k )
11
(b) 220
20
55
(d) 110
[2015]
(c)
18
14. Let a, b, c, d and e be distinct positive number. If a,
1 1 1
b, c and , , both are in A.P. and b, c, d are in
c d e
G.P. then
(a)
8.38
Complete Mathematics—JEE Main
(a) a, b, c are in G.P
(c) a, c, e are in A.P.
(b) a, b, c are in A.P.
(d) a, c, e are in G.P.
2.
6.
10.
14.
(a)
(b)
(b)
(b)
3.
7.
11.
15.
(c)
(d)
(a)
(d)
4. (a)
8. (c)
12. (c)
Level 1
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
64.
(c)
(c)
(a)
(c)
(c)
(b)
(d)
(a)
(d)
(c)
(a)
(a)
(a)
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
57.
61.
65.
(c)
(d)
(a)
(b)
(c)
(a)
(a)
(a)
(d)
(a)
(b)
(a)
(c)
43. (b)
44. (b)
45. (b)
46. (a)
47. (b)
48. (b)
Previous Years' B-Architecture Entrance
Examination Questions
Concept-based
(a)
(d)
(b)
(a)
42. (b)
49. (a)
Answers
1.
5.
9.
13.
41. (d)
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
62.
66.
(a)
(d)
(a)
(d)
(c)
(d)
(a)
(a)
(a)
(c)
(a)
(a)
(d)
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
63.
67.
(b)
(b)
(c)
(c)
(d)
(b)
(b)
(c)
(b)
(c)
(b)
(b)
(a)
1. (d)
2. (b)
3. (d)
4. (a)
5. (d)
6. (a)
7. (c)
8. (b)
9. (b)
10. (a)
11. (b)
12. (c)
13. (b)
14. (d)
Hints and Solutions
Concept-based
1. a1, a2, a3, . . . is a G.P.
fi log5 a1, log5 a2, log5 a3, . . . are in A.P.
2. If r > 1 is the common ratio of the G.P., then
1
> 2.
r
and d = an – an – 1.
b
+
r
b + br = xb fi x – 1 = r +
3. Use : an = Sn – Sn – 1
4. Middle terms are nth and (n + 1)th terms.
68. (c)
69. (b)
70. (b)
71. (a)
72. (c)
73. (d)
74. (c)
75. (d)
a – b = tn = a + (n – 1)d
76. (d)
77. (d)
78. (a)
79. (b)
and a + b = tn + 1 = a + nd
80. (c)
81. (d)
82. (a)
83. (d)
84. (b)
85. (c)
2n
[2a + (2n – 1)d] = 2na
2
5. Use : 3 + 7(n – 1) = 63 + 2(n – 1)
S2n =
Level 2
86. (c)
87. (a)
88. (a)
89. (c)
90. (b)
91. (c)
92. (b)
93. (a)
94. (a)
95. (a)
96. (b)
Previous Years' AIEEE/JEE Main Questions
6. For example, if a = 2, b =
b + 1, c forms a G.P.
2 – 1, c = 1, then a,
7. alog x = blog y = clog z = k (say)
fi a = k1/log x, b = k1/log y, c = k1/log z
Now, a, b, c are in G.P.
1. (b)
2. (b)
3. (b)
4. (a)
5. (b)
6. (c)
7. (d)
8. (c)
9. (b)
10. (c)
11. (a)
12. (a)
13. (a)
14. (d)
15. (b)
16. (d)
fi log x, log y, log z are in H.P.
17. (d)
18. (a)
19. (b)
20. (c)
However, nothing can be said about x, y, z.
21. (a)
22. (d)
23. (b)
24. (b)
25. (c)
26. (b)
27. (a)
28. (c)
29. (c)
30. (c)
31. (c)
32. (c)
33. (b)
34. (c)
35. (a)
36. (c)
37. (b)
38. (a)
39. (d)
40. (b)
fi
1
1
1
are in A.P.
,
,
log x log y log z
8. S3 =
a (1 - r 3 )
a (1 - r 6 )
, S6 =
1- r
1- r
S3
1 - r 3 125
125
=
=
fi
S6
1 - r 6 152
152
Progressions
fi 1 + r3 =
152
27 Ê 3 ˆ 3
fi r3 =
=
125
125 Ë 5 ¯
1 Ê 1 3ˆ
3a + b
+ ¯ =
Ë
4 a b
4ab
A3 =
a, a – 18, a – 9, a
1
15
(3a + b) (a + b) (a + 3b) =
64
2
1
5
and
(a + 3b) (a + b) (3a + b) =
3
18
64(ab)
As a, a – 18, a – 9 are in G.P.
Thus, (ab)3 =
(a – 18)2 = a(a – 9)
fi ab = 3.
fi r = 3/5
Now,
9. As the common difference is 9, we let the four numbers be
2
2
fi a – 36a + 324 = a – 9a
a+b
b+c
=
–b
1 - ab
1 - bc
13. b –
fi 27a = 324 fi a = 12
Thus, sum of the four numbers is 21
b - ab2 - a - b
b + c - b + b2 c
=
1 - ab
1 - bc
fi
10. Let two positive numbers be a, b. Then
a + b = 2(7) = 14 and
ab = 5 or ab = 25
fi–
Now, a2 + b2 = (a + b)2 – 2ab = 196 – 50 = 146
8 (1 - r )3
1 - r3
= 24 fi
1 - 2r + r 2
1 + r + r2
fi 2abc = a + c
2b =
\
=3
fi 2r2 + 5r + 2 = 0 fi r = –2, –1/2
1
1
, b,
are in A.P.
a
c
S=
Now, t5 = ar4 = 3/16
12. Let three arithmetic means between a and b be a1,
a2, a3. Then a, a1, a2, a3, b are in A.P. Let common
difference of this A.P. be d. Then
1
d = (b – a)
4
1
a1 = a + b = (3a + b)
4
1
a2 = a + 2d = (a + b)
2
1
a3 = a + 3d = (a + 3b)
4
If A1, A2, A3 are arithmetic means between 1/a and
1/b, then
1 Ê 3 1ˆ
a + 3b
+ ¯ =
A1 =
Ë
4 a b
4ab
1 1
+
a c
14. Let
As |r| < 1, r = –1/2, a = 3.
1 Ê 1 1ˆ
a+b
A2 =
+ ¯ =
Ë
2ab
2 a b
c (1 + b2 )
a (b2 + 1)
=
1 - bc
1 - ab
fi – a + abc = c – abc
1 2
(a + b2) = 73.
2
a
a3
= 2,
= 24
11.
1- r
1 - r3
fi
\
15 2
= 27
5 18
5
55
555
+
+
+...
17 (17)2 (17)3
1
5
55
+
+ ...
S =
2
17
(17)
(17)3
Subtracting, we get
1 ˆ = 5 + 50 + 500 + 5000 + . . .
Ê
1
Ë 17 ¯ S
17 (17)2 (17)3 (17)4
16
5 17
5
85
S =
=
fiS=
17
1 - 10 17
7
112
2016
15.
 (-1)r
r =1
2016
But
 (-1)r
r =1
2016
2016
r =1
r =1
(a + rd) = a  ( -1)r + d  ( -1)r (r )
2016
= 0 and
 (-1)r (r )
= 1008
r =1
Level 1
16. 2(2a + b) = (5a – b) + (a + 2b) fi b = 2a.
Also, (a – 1)2 (b + 1)2 = (ab + 1)2
fi (a – 1) (b + 1) = ± (ab + 1)
fi (a – 1) (2a + 1) = ± (2a2 + 1)
8.39
Complete Mathematics—JEE Main
8.40
2a2 – a – 1 = 2a2 + 1
or 2a2 – a – 1 = – 2a2 – 1
fi a = – 2, a = 1/4
fi t = 4, 48.
\ g = 4, l = 48
Let a = 4a1, b = 4b1, where hcf (a1, b1) = 1.
1
1
1
1
+ +
=0
17. +
a c-b c a-b
fi
Also, 4a1 b1 = 48 fi a1 b1 = 12.
fi a1 = 1, b1 = 12 or a1 = 3, b1 = 4
a+c-b
a+c-b
+
=0
a (c - b )
c( a - b )
Thus, smaller of a, b, is 4 or 12.
99 2 2
2
21.
< +
+ . . . + n -1
100 3 32
3
As a + c – b π 0, a(c – b) + c(a – b) = 0
fib=
2ac
fi a, b, c are in H.P.
a+c
18. 0 = Sp =
Sp + q =
=
p
2
[2a + (p – 1)d ] fi d =
1- p
2
p+q
[2a + (p + q – 1)d ]
2
1
(p + q) [2a + (p – 1)d + qd ]
2
fi ma + nb =
ab
22. a, b, c, d are in A.P.
(1)
n
Thus, n = k a =
4 gl
=
fi (g + l)2 =
H.M. =
2 ab
a+ b
2a b
a+ b
169
48
169
(4) (192) = 132 (4)2
48
fi g + l = 52
Note that g, l are roots of
23. a + b =
(2)
20. We have gl = ab = 192.
2gl
1
= 169 : 48
Also, (g + l ):
g+l
2
( g + l )2
1 1 1 1
, , , are in H.P.
a b c d
1 1 1 1
, , , are in H.P.
fi
d c b a
Now multiply each term by abcd
ab (m + n)
=
= k(say)
b
a
From (1) and (2)
1
k ( a + b ) = ab fi k =
2
fi
1
1
> n - 1 fi 3n – 1 > 100
100 3
24.
or t – 52t + 192 = 0
4+ 2
5+ 2
, ab =
8+2 2
5+ 2
16 + 4 2
2a b
=
=4
a +b
4+ 2
2
1
1
+
=
cos x
cos ( x - y) cos ( x + y)
fi 2(cos2x – sin2y) = 2cosx cosx cosy.
fi cos2x(1 – cosy) = sin2y = 1 – cos2y
y
fi cos2x = 1 + cosy = 2cos2 Ê ˆ .
Ë2 ¯
y
fi cos x sec Ê ˆ = ± 2 .
Ë2 ¯
25. a(2n) = 1 +
>1+
1 1 1
1
+ ... + 2n
+ +
2 3 4
2 –1
1 Ê 1 1ˆ
1 1 1 1
+
+ Ê + + + ˆ + ...
+
Ë 5 6 7 8¯
2 Ë 3 4¯
1
1 ˆ
1
Ê
+ ... + 2 n ˜ –
+ Á 2n – 1
Ë2
¯
+1
2
22 n
t2 – (g + l)t + gl = 0
2
100
Thus, least value of n is 7.
fi m a ( a - b) = n b ( a - b)
fi
fi
1-1 3
fi
ma + nb
1
= (m + n) =
2
m+n
m
(2 3) ÈÎ1 - (1 3)n - 1 ˘˚
99
>
fin–1>5fin>6
( p + q)q
=
1- p
19.
fi
>
1 1 2 4
22 n - 1
n
1
1
+ + + + ... + 2 n + ÊÁ - 2 n ˆ˜ > if n > 1
Ë2 2 ¯ 2
2 2 4 8
2
Progressions
\ a(200) > 100.
4
44 444
+
+
+ ...
26. S =
19 192 193
=1+
(1)
1
4
44
S = 2 + 3 + ...
19
19
19
Subtracting (2) from (1), we get
Ê 18 ˆ S = 4 + 40 + 400 + ...
Ë 19 ¯
19 192 193
=
fi
S=
fi
(2)
x
y
fi 164 = 3m + 5m – [3(m – 1)2 + 5(m – 1)
= 6m + 2
fi
m = 27.
b
34. G1 = ar, G2 = ar 2 where r = Ê ˆ
Ë a¯
19 4
38
¥ =
18 9
81
35. Use A1 + A2 = a + b and G1 G2 = ab.
36. Let n be the number of terms. Then
n
2s
.
s = (a + l ) fi n =
2
a+l
Also,
u
log a log b log c log d
=
=
=
1/ z
1/ u
1/ x
1/ y
Now, use log a, log b, log c, log u are in A.P.
fi
d=
37. x = 2 S
29. (a2p2 – 2abp + b2) + (b2p2 – 2bcp + c2) + (c2p2 – 2cdp
+ d 2) £ 0
where S =
14
1
1 1 1
1
S=
+
+ ... =
+ +
1–1 2
2
4 8 16 32
ap – b = 0, bp – c = 0, cp – d = 0
30. a, b, c are in A.P.
fi2
ax + 1
,2
bx + 1
,2
=
2ax, 2bx, 2cx are in G.P.
cx + 1
fi
\
are in G.P.
357
357
+
+ ...
31. 2 ◊ 357 = 2 +
1000 (1000)2
=2+
38.
357 1000
357
2355
=2+
=
999
999
1 – 1 1000
4 7 10
+
+
+ ...
5 52 53
1
1 4
7
S = + 2 + 3 + ...
5
5 5
5
Subtracting (2) from (1), we get
4
3 3
3
35
S = 1 + + 2 + 3 + ... = 1 +
5
5 5
1
–
15
5
32. S = 1 +
1
1
¥2=
4
2
S = 1.
x=2
a, b, c, d are in G.P.
fi
1 2 3
4
+ +
+
+ ...
4 8 16 32
fi
As a, b, c, d and p are real this is possible if and
only if
fi
l – a (l – a ) (l + a )
l 2 – a2
=
=
2 s – (l + a)
n – 1 2 s – (l + a)
1
1 2
3
+
+ ...
S= +
2
8 16 32
Subtracting, we get
fi (ap – b)2 + (bp – c)2 + (cp – d)2 £ 0
b c d
= =
a b c
.
= a + ar 3 = a + b = 2A
fi x log a = y log b = z log c = u log d
fi
13
G12 G2 2
G 3 + G23
a 3r 3 + a 3r 6
+
= 1
=
G2
G1
G1 G2
a2 r 3
log a, log b, log c º are in A.P.
28. a = b = c = d
– 1
2
4 19
4
=
9
1 – 10 19
z
35
.
16
33. Tm = Sm – Sm
27. a1/x = b1/y = c1/z = ... = k(say)
log a log b log c
=
=
= º log k,
fi
x
y
z
Now, use
S=
3 7
=
4 4
a
a2
= 20 and
= 100.
1– r
1 – r2
fi
(1)
(2)
400 (1 - r )2
1 - r2
= 100
1– r
1
=
fi 4 – 4r = 1 + r
1+ r
4
fi r = 3/5
fi
Thus,
a
= 20 fi a = 8
25
39. ar 7 = 128 = 27 and r = 2 fi a = 1.
8.41
8.42
Complete Mathematics—JEE Main
Product of first five terms = a5 r10 = 1024.
40. l = ARp
– 1
, m = ARq
– 1
a + ( p - 1)d
a + (q - 1)d
a + (r - 1)d
D=
, n = ARr
– 1
p
q
r
fi (a2 + c2 + 4ac) (a2 + c2 – 2ac) = 0
1
1
1
where a = log A, d = log R.
Use C1 Æ C1 – (a – d)C3 – dC3 to show D = 0.
p
a+
[2a + ( p - 1)d ]
p2
2
41.
= 2 fi
q
q
a+
[2a + (q - 1)d ]
2
Now, put p = 11 and q = 41.
42. a =
2, r =
p –1
d
p
2
=
q –1
q
d
2
fi [(a + c)2 + 2ac] (a – c)2 = 0
fi (a + c)2 + 2ac = 0
2 (35 - 1)
3 -1
2 (242) ( 3 + 1)
= 121 ( 6 + 2 )
( 3 - 1) ( 3 + 1)
1
43.
n(n + 1) (2n + 1) = 1015
6
fi (2n) (2n + 1) (2n + 2) = (28) (29) (30)
1
44. tr = 1 + 2 + ... + r = r (r + 1) (2r + 1)
6
1
= (2r3 + 3r2 + r)
6
2
2
n
1 1
\ Â tr = ◊ n2(n + 1)2
3 4
r =1
1 1
1
n(n + 1)
+ ◊ n(n + 1) (2n + 1) +
2 6
12
=
1
n(n + 1)[n2 + n + 2n + 1 + 1]
12
=
1
n(n + 1) (n2 + 3n + 2)
12
=
1
n(n + 1)2 (n + 2)
12
22
1
(22) (23)2 (24) = 23276
\ Â tr =
12
r =1
45. 2b = a + c, b2 =
fi
(a + c)
4
2
2a2 c2
a2 + c2
2
=
a π c]
fi 4b + 2ac = 0
a
, b, c are in G.P.
2
46. am = n, an = m
fi –
fi (m – n)d = an – an = – (m – n)
fi d = – 1.
Now, ar – am = (r – m) (– 1)
47. As b2 = ac, ax2 + 2bx + c = 0 can be written as
ax2 + 2 ac x + c = 0 fi ( a x + c )2 = 0
fix= –
2a c
2
a2 + c2
c
,–
a
c
.
a
c
is a root of dx2 + 2ex + f = 0
a
\–
fi 2n = 28 or n = 14.
2
[∵
2
fi ar = am – r + m = m + n – r.
3
a (r10 - 1)
S10 =
=
r -1
=
fi (a2 + c2)2 + 2ac(a2 + c2) – 8a2c2 = 0
c
c
fi d Ê ˆ – 2e
+ f = 0.
Ë a¯
a
fi
1
d
f
– 2e
+
= 0.
a
ac c
fi2
e
d f
+
=
b
a c
fi
d e f
, , are in A.P.
a b c
48. 2, 2 x + 1, 2 x + 3 are in G.P.
fi (2 x + 1)2 = 2(2x + 3) fi 22x = 5
fix=
1
log25
2
1
x, |x + 1|, |x – 1| are in A.P.
2
1
2|x + 1| = x + |x – 1|
2
49. As
(1)
If x < – 1, then (1) give us
1
– 2(x + 1) = x – (x – 1)
2
fix=–2
In this case common difference is 2 and sum to 10
terms is
10
[2(– 1) + (9) (2)] = 85
2
Progressions
If – 1 £ x < 1,
1
2(x + 1) =
x – (x – 1) fi x = – 2/5
2
In this case common difference is 4/5, and sum to 10
terms is
10 È Ê 1 ˆ
Ê 4ˆ ˘
2 - + (9) Ë ¯ ˙ = 34
5 ˚
2 ÍÎ Ë 5 ¯
If x ≥ 1, then (1) becomes
1
x+x–1
2x + 2 =
2
1
x=–3
fi
2
Not possible as x ≥ 1.
50.
2ab
: ab = 4 : 5 fi
a+b
fi
fi
fi
a+b
ab
3
3
2
1 + 2 + ... + r
r (r + 1)
=
1 + 3 + ... (2r - 1)
4r 2
16
 tr
r =1
But (x – a) (x – b) is least when x =
\ T1 T2 T7 is least when d =
2
=
(r + 1)
4
2
1 Ê 17 2 ˆ
= Á Â r – 1˜
4 Ër =1
¯
1 Ê1
ˆ
(17) (18) (35) - 1¯ = 446
Ë
4 6
52. Given equation implies
t 81
= 28 where t = 32sin2q
+
3 t
fi t2 – 84t + 243 = 0 fi t = 3, 81
But 1/9 £ t £ 9. Thus, t = 3.
\ a1 = 1, d = 13
fi
a5 = a1 + 4d = 53.
56. Write the given series as
Ê 1 ˆ + Ê 1 + 1 ˆ + Ê 1 + 1 ˆ ... + Ê 1 + 1 ˆ
˜
ÁË
˜
˜ Á
Ë1 + 2 ¯ ÁË
22 ¯ Ë
23 ¯
2n ¯
fi S2m
2ac
fi a, b, c are in H.P.
a+c
53. Required sum
= (2 + 4 + ... + 100) + (5 + 10 + ... + 100) –
(10 + 20 + ... + 100)
50
20
10
(2 + 100) + (5 + 100) – (10 + 100)
=
2
2
2
= (50) (51) + (10) (105) – 550
= 2550 + 1050 – 550 = 3050
54. T1 T2 T7 = (T7 – 6d) (T7 – 5d)T7
= 9(9 – 6d) (9 – 5d)
= m(2m + 1)2 – 2(2m)2 = m(2m – 1)2
1
ˆ
Ê r +1– r ˆ
Ê
58. ar = tan–1 Á
= tan–1 Á
Ë 1 + (r + 1)r ˜¯
Ë 1 + r (r + 1) ˜¯
= tan–1(r + 1) – tan–1(r)
n
Thus, Sn =
 ar
= tan–1(n + 1) – tan–1(1)
r =1
Ê n +1–1 ˆ
Ê n ˆ
= tan–1 Á
= tan–1 Á
Ë 1 + (n + 1) (1) ˜¯
Ë n + 2 ˜¯
fi (a + c)2 – (a – c)2 = 2b(a + c)
4ac = 2b(a + c)
– 1
2m (2m + 1)2
= m(2m + 1)2
2
Now, put n = 2m – 1.
(a + c) (a + c – 2b) = (a – c)2
fib=
1 Ê 3 9ˆ
33
+ ¯ =
Ë
2 2 5
20
55. We have
57. We have S2m =
=
fi
1
(a + b).
2
1Ê
1
1- nˆ
Ë
2
2 ¯ =n+1– 1
=n+
1–1 2
2n
a
b 5
a
1
fi
+
=
= 2,
b
a 2
b
2
a : b = 4 : 1 or 1 : 4
Thus,
9
3
= 729 Êd – ˆ Ê d – ˆ
Ë
¯
Ë
5¯
2
5
2
=
3
51. t r =
2 ab
4
=
a+b
5
8.43
59. tr=
1
1
1
1
–
+
=
r (r + 1) (r + 2)
2r r + 1 2(r + 2)
[split into partial fractions]
1 ˆ 1Ê 1
1 ˆ
1 Ê1
–
= Á –
– Á
˜
˜
2 Ë r r + 1¯ 2 Ë r + 1 r + 2 ¯
n
fi
Sn =
 tr
r =1
=
Thus, S30 =
=
1 ˆ 1 Ê1
1 ˆ
1Ê
ÁË 1 – n + 1˜¯ – ÁË 2 – n + 2 ˜¯
2
2
1 n
n
–
2 n + 1 4 ( n + 2)
1 30
30
30 È 64 – 31 ˘
495
–
=
=
Í
˙
2 31 4(32)
1984
4 Î (31) (32) ˚
60. 9(n!)2 = n!(n+ 1)!
fi
9=n+1fin=8
This shows that 8!, 5(8!), 9(8!) are in A.P.
8.44
Complete Mathematics—JEE Main
2
.
3
= ar n
61. Here a = 81, r =
Also, Sn – Sn
– 1
65. Coefficient of x99
– 1
= – (1 + 3 + 5 + ... 199) = – 1002 = – 10000
.
y = (Sn – Sn
+ 3 (Sn
= ar
n – 1
= arn
– 4
– 1)
– 2
– 3(Sn
– Sn
– 3ar
– 1
– 3)
n – 2
– Sn
– (Sn
+ 3ar
n
– 2)
– 3
n – 3
– Sn
– ar
 tr
– 4)
(r3 – 3r 2 + 3r – 1)
67. x =
=
Â
r +
r =1
1
n(n + 1) (n + 2)
6
y – 1 yx
xy – x
=
◊
y–x
y–x
y
68.
 [(2r – 1) (2r ) – (2r )2 ]
fi
x + yz = x(y + z).
a
a3
108
= 3,
=
where | r | < 1.
3
1– r
13
1– r
r =1
1
= (2m – 1) (2m) (4m – 1) – 2
6
m –1
Â
\
r
27(1 - r )3
1- r
3
=
1
m(2m – 1) (4m – 1) – (m – 1)m
3
=
1 Ê n + 1ˆ
n + 1ˆ Ê n - 1ˆ
n (2n + 1) - Ê
Ë
¯
Ë
3 2
2 ¯Ë 2 ¯
fi
1 – 2r + r 2
1 + r + r2
Â
[(r + 1)log a – (r – 1)log b]
r =1
=
n
n
2
L = lim (1 – 3–2) (1 + 3–2) (1 + 3–4) ... (1 + 3 –2 )
nƕ
3
n
= lim (1 - 3-2 ) (1 + 3–4) º (1 + 3 –2 )
nƕ
) =1
L = 3/2
64. Note that
1
1
1
+
+ ... +
a1 a2
an
1 1
1 1
ˆ
< ÊÁ 1 + + 2 + ...ˆ˜ ÊÁ 1 + + 2 + ...˜
¯
Ë
¯
Ë
3 3
5 5
15
Ê 1 ˆÊ 1 ˆ
= Á
=
<2
˜
Á
˜
Ë 1 – 1 3¯ Ë 1 – 1 5¯
8
[∵ | r | < 1]
n
69. S =
(1 + 3–4) ... (1 + 3–2 )
nƕ
4
13
fi (3r – 1) (r – 3) = 0 fi r = 1 3
nƕ
2
=
fi 9r2 – 30r + 9 = 0 fi 3r2 – 10r + 3 = 0.
63. (1 – 3–1)L = lim (1 – 3–1) (1 + 3–1) (1 + 3–2)
n +1
108
13
fi 13 – 26r + 13r2 = 4 + 4r + 4r2
1
(n + 1) [2n(2n + 1) – 3(n – 1)]
12
1
=
(n + 1) (4n2 – n + 3)
12
= ... = lim (1 - 3-2
=
r =1
=
fi
È1
ÍÎ 6
fi yz – xz = xy – x
m –1
2
=
z=
= 12 + (1) (2) + 32 + (3) (4) + ...
(2m – 3) (2m – 2) + (2m – 1)2
2m – 1
1
2
1
1
1
b
,y=
,z=
=
b–a
1– a
1– b
1– a b
62. Let n = 2m – 1, then
– 1
=
r =1
n – 4
2
Thus, x = r = .
3
S2m
1
1
r(r + 1) = (r 2 + r)
2
2
1
˘
n(n + 1) (2n + 1) + n(n + 1)˙
˚
2
66. tr = 1 + 2 + ... + r =
Let y = numerator of x, then
n
[(2 + n + 1)log a – (0 + n – 1)log b]
2
n
Ê an + 3 ˆ
log Á
Ë b n – 1 ˜¯
2
70. t5 = ar 4 = 2.
=
We have t1t2 ... t9 = a9 r 1
+ 2 + ... + 8
= a9 r 36 = (ar4 ) 9 = 29 = 512
71. s =
a
a
a
fi1–r=
fir=1–
1–r
s
s
Sn =
a(1 - r n )
a
=
1- r
a s
È Ê
aˆ n ˘
Í1 – Ë 1 – ¯ ˙
Î
s ˚
aˆ n ˘
˙
s¯ ˚
72. Note that < an > is a G.P. with first term equal to 5
and common ratio 1/2.
È
= s Í1 – Ê 1 –
Î Ë
Progressions
•
 a2 n – 1
\
=
n =1
73.
74.
a1
=
1 – r2
2ab
: ab = 12 : 13
a+b
5
= 20/3.
1–1 4
fi
fi
fi
a
2
3
=
or
fi a : b = 4 : 9 or 9 : 4
b
3
2
5
ab, a + b + ab = 26
8
13
ab = 26 fi ab = 16 and a + b = 10
8
Thus, numbers are 2 and 8.
\
=
+ a 2n – 2
= (an – an – 1) (an + an – 1)
– 1
1
1– r
a
,B=
fi (a2 + a25) + (a6 + a21) + (a9 + a18) = 165
fi 3(a1 + a26) = 165
\
a1 + a2 + ... + a26
80. We are given Sn = cn2. Now,
tn = Sn – Sn – 1 = cn2 – c(n – 1)2 = (2n – 1)c
We have
n
t12 + t22 + ... + tn2 =
n
a
log (1 - A )
=
b
log (1 - B -1 )
77. Note that ak – ak –1 = ak – 1 – ak – 2 " k ≥ 3 implies
a1, a2, a3 ..., a12 are in A.P.
Also, d = – 4. We have
12S = (a1 – a2) (a1 + a2) + (a3 – a4) (a3 + a4)
+ ... + (a11 – a12) (a11 + a12)
= 4(a1 + a2 + ... + a12)
12
=4¥
[2a1 + (12 – 1) (– 4)]
2
S = 2[2(45) – 44] = 92
78. We have
1 – (– x )2013 1 + x 2013
P(x) =
=
1 – (– x )
1+ x
 (4k 2 – 4k + 1)
= c2
k =1
1 – rb
-1
 c2 (2k - 1)2
k =1
1
b log r = log(1 – 1/B)
fi a1 + a26 = 55
26
=
(a1 + a26)
2
= 13 ¥ 55 = 715
4
È4
˘
= c2 Í n(n + 1)(2n + 1) - n(n + 1) + n ˙
2
Î6
˚
1
1
fi 1–r =
, 1 – rb =
A
B
fi a log r = log(1 – 1/A) and
fi
1 2013
(3
+ 1)
4
a2 + a6 + a9 + a18 + a21 + a25 = 165
a
\
[∵y = 1 = x – 2]
Thus,
– (an – 1 – an – 2) (an – 1 + an – 2)
= d(an + an – 1 – an – 1 – an – 2) = d(2d) = 2d2
76. A =
= Q(1) = P(3)
a1 + a26 = a2 + a25 = a3 + a24 = ... = a9 + a18.
2ab 16
=
, a + b + ab = 26
a+b 5
75. a n2 – 2a2n
 ai
i=0
79. We have
b 13
=
a 6
a+b=
2012
Also
2 ab 12
=
a + b 13
a
+
b
fi
8.45
=
c2
n È2 2 n2 + 3n + 1 – 6 n – 6 + 3˘˚
3 Î
(
)
1
n 4n2 – 1 c 2
3
81. Sum to n terms of an A.P. is of the form an + bn2
and not of the form c + bn + an2 where c π 0.
(
=
82. |2x – 1| < 1
fi
fi
)
– 1 < 2x – 1 < 1
0<x<1
83. As a + c > 0, we must have
(a + c)2 + 4b2 – 4(a + c)b < 0
fi (2b – a – c)2 < 0
This is not possible as 2b Рa Рc ΠR.
84. (a2 + b 2)2 = (a + b) (a3 + b3)
fi a 4 + b 4 + 2a 2b 2 = a 4 + ab3 + a 3b + b4
fi ab(a – b) 2 = 0
fi ab [(a + b)2 – 4ab] = 0
8.46
Complete Mathematics—JEE Main
fi
c È 4b 2 4c ˘
– ˙ =0
Í
a Î a2
a˚
fi
Next, a, ab, b are in A.P.
fi –
2b 2c
=
a
a
fi
1
1 1
88. a ÊÁ + ˆ˜ , b ÊÁ +
Ëc
Ë b c¯
cD = 0
fi
a + b = 2ab
b+c=0
85. Let a1 = a, a2 = ar, a3 = ar 2 and a4 = ar3.
As a1, a2, a3, a4 are distinct, r π 1
1 1
1
fi a ÊÁ + ˆ˜ + 1, b ÊÁ +
Ë b c¯
Ëc
A.P.
Ê bc + ca + ab ˆ
bc + ca + ab ˆ
fi aÁ
, b ÊÁ
˜¯ ,
˜
Ë
¯
Ë
abc
abc
bc + ca + ab ˆ
˜¯ are in A.P.
abc
fi a, b, c are in A.P.
b3 = a(1 + r + r 2)
and b4 = a(1 + r + r 2 + r 3).
89. sin d [cosec a1 cosec a2 + cosec a2 cosec a3 +
+
cosec an – 1 cosec an]
sin (an - an - 1 )
sin (a3 - a2 )
sin (a2 - a1 )
=
+
+ +
sin a1 sin a2
sin a2 sin a3
sin an -1 sin an
As b2 – b1 = ar π ar2 = b3 – b2.
fi b1, b2, b3, b4 are not in A.P.
1 + r + r 2 b3
b2
=1+rπ
=
b2
1+ r
b1
= (cot a1 – cot a2) + (cot a2 – cot a3)
+
fi b1, b2, b3, b4 are not in G.P.
(
Since
+ (cot an – 1 – cot an)
= cot a1 – cot an
3 5
2n - 1
90. 1 + + +
+
2 3
n
–r
1
1
– =
b2 b1 a (1 + r )
1
– r2
1
–
=
b3 b2 a (1 + r ) 1 + r + r 2
1ˆ
Ê 1 1ˆ
˜¯ +1, c ÁË + ˜¯ + 1 are in
a
a b
c ÊÁ
Ë
b1 = a, b2 = a + ar = a(1 + r),
Next,
1ˆ
Ê 1 1ˆ
˜¯ , c ÁË + ˜¯ are in A.P.
a
a b
1
1
= (2 – 1) + ÊÁ 2 - ˆ˜ + ÊÁ 2 - ˆ˜ +
Ë
2¯ Ë
3¯
)
1
1
1
1
– π
–
we get b1, b2, b3, b4 cannot
b2 b1 b3 b2
be in H.P.
Level 2
= 2n – Hn
91. a, b, c are in H.P.
1 1 1
fi , , are in A.P.
a b c
fi
86. The given sequence can be written as
5 20 10 20
, , , ,
2 13 9 23
The reciprocal of the numbers in the sequence are
8 13 18 23
, , , ,
20 20 20 20
This is an A.P.
nth term of this A.P. is
8 + 5 (n - 1)
3 + 5n
=
an =
20
20
Thus, nth term of the given sequence is
87. H.M. between a and b is
2ab
2 (1 / 28) (1 / 10)
1
=
=
a+b
1 / 28 + 1 / 10
19
1
+ ÊÁ 2 - ˆ˜
Ë
n¯
a+b+c
a+b+c
a+b+c
–2,
–2,
– 2 are in
b
c
a
A.P.
fi
b+c-a c+a-b a+b-c
,
,
are in A.P.
b
c
a
fi
a
b
c
,
,
are in H.P.
b +c - a c +a - b a +b- c
92. Last term of the nth row
1
n (n + 1)
2
Note that sum of the terms in the nth row
=1+2+3+
20
.
5n + 3
+n=
= sum to n terms of an A.P. whose first term in
1
n(n + 1) and common difference is (–1).
2
1
1
1
= n ÈÍ2 ÊÁ ˆ˜ n (n + 1) + (n - 1) (- 1) ˘˙ = n (n 2 + 1)
2 Î Ë 2¯
˚ 2
Progressions
96. a + 8b + 7c = 0
93. We have
1
2a + b
A1 = a + (b – a) =
3
3
9a + 2b + 3c = 0
a + 2b
and A2 =
3
as b = 6, we get a = 1, c = – 7
5a + 5b + 5c = 0
The roots of quadratic equation
\ A1 + A2 = a + b
1
1 2 1
1 2b + a ˆ
= ÊÁ + ˆ˜ = ÊÁ
Also,
˜
¯
Ë
H1
3 a b
3 Ë ab ¯
and
x2 + 6x – 7 = 0 fi x = 1, – 7
Now,
1
1 2a + b ˆ
= ÊÁ
˜
H2
3 Ë ab ¯
Also,
n
•
=
6
 ÊÁË 7 ˆ˜¯
n=0
n
=
1
=7
1- 6/7
1. t5 = 2. Let r be the common ratio of the G.P.
t 1t 2t 3t 4t 5t 6t 7t 8t 9
Therefore,
A1 + A2
ab
ab (a + 2b) (2a + b)
=
=
H1 + H 2
H1 H 2
9
(ab) 2
=
Ê 2 ˆ Ê 2 ˆ Ê 2 ˆ Ê 2 ˆ (2)(2r )(2r 2 )(2r 3 )(2r 4 )
Ë r 4 ¯ Ë r3 ¯ Ë r 2 ¯ Ë r ¯
= 29 = 512.
2. Let E = 21/441/881/16 … = 25
where
2 (a 2 + b 2 ) + 5 ab
=
9 ab
1 2 3
4
+ + + +
4 8 16 32
1
1 2
3
fi
S= + + +
2
8 16 32
Subtracting, we get
S=
1
+
2n
1
1
1
= ÊÁ1 + + + ˆ˜ – ÊÁ1 + + +
Ë
¯
Ë
2
2
2n
Ê 1 1ˆ
 ÁË a + b ˜¯
n=0
Previous Year’s AIEEE/JEE Main Questions
a+b
H1 H2
ab
1
1
+
94.
+
n +1 n + 2
6
1
1 1
= 1=
+
7
7
a b
•
Thus,
1
1
1 1
=
+
+
H1 H 2
a b
fi H1 + H2 =
8.47
1ˆ
˜
n¯
1 1 1
1
1ˆ
Ê
= Á1 - + - + +
- ˜
Ë
2 3 4
2n - 1 2n ¯
1 1
1
1
+ 2 ÊÁ + + + ˆ˜ – ÊÁ1 + + +
¯
Ë2 4
Ë
2n
2
= a (2n)
1ˆ
˜
n¯
95. Let m be the minimum number filled in the boxes.
Let a, b, c, d be the numbers in the boxes sharing a
side with the box containing m, then
1
(a + b + c + d)
m=
4
fi (a – m) + (b – m) + (c – m) + (d – m) = 0
(1)
As a – m ≥ 0, b – m ≥ 0 , c – m ≥ 0, d – m ≥ 0,
1
1 1 1
1
S= + + + +
2
4 8 16 32
1
1
= 41 =
1- 2 2
fiS=1
\ E = 21 = 2
3. 1, log9 (31 – x + 2), log3 (4(3x) – 1) are in A.P.
log3 (31- x + 2)
, log3 (4(3x) – 1) are in A.P.
log3 9
fi 2, log3 (31 – x + 2), 2 log3 (4(3x) – 1) are in A.P.
fi 1,
fi 32, 31 – x + 2, (4(3x) – 1)2 are in G.P.
fi (31 – x + 2)2 = 32 (4(3x) – 1)2
(1) is possible if and only if
3
x
+ 2 = 3(4(y) – 1) where y = 3
y
a – m = 0, b – m = 0, c – m = 0 d – m = 0
fi
Thus, each of the box contains m.
fi 12y2 – 5y – 3 = 0
As the first box contains 5, m = 5.
fi (4y – 3) (3y + 1) = 0
Therefore, x = 5
fi y = 3/4
[
3y + 1 > 0]
8.48
Complete Mathematics—JEE Main
fi 3x = 3/4 fi x = 1 – log34
3
3
3
3
3
4. 1 – 2 + 3 – 4 +
–8 +9
= (13 – 23) + (33 – 43) +
3
+ (73 – 83) + 93
= (1 – 2) (12 + 22 + 2) + (3 – 4) (32 + 42 + (3)(4))
+ (5 – 6) (52 + 62 + (5) (6))
+ (7 – 8) (72 + 82 + (7)(8)) + 93
= – (12 + 22 + 32 +
8 2)
– (2 + 12 + 30 + 56) + 93
1
(8)(9)(17) – 100
6
= 729 – 204 – 100 = 425
= 93 –
5. Let S1 = {2x | 1 £ x £ 50}
S2 = {5x | 1 £ x £ 20}
S1 « S2 = {10x | 1 £ x £ 10}
12 + 2 ◊ 22 + 32 + 2 ◊ 42 + 52 + 2 ◊ 62 + … +
(2m – 1)2 + 2(2m)2
1
(2m)(2m + 1)2
2
fi 12 + 2 ◊ 22 + 32 + 2 ◊ 42 +
=
x ŒS2
= m(4m2 + 4m + 1) – 8m2
x ŒS1nS2
Ê 50 ˆ (51) + 5 Ê 20 ˆ (21) - 10 Ê 10 ˆ (11)
= 2
Ë 2¯
Ë 2¯
Ë 2¯
= (50) (51) + 5(10) (21) – 5(10) (11)
= (50) [51 + 21 – 11] = 3050
a ˆ
fi 20 Ê
= 100
Ë1+ r ¯
fi
=
1
1
1
,y=
,z=
1- b
1- c
1- a
a, b, c are in A.P.
10. We have x =
1
1
1
are in H.P.
,
,
1- a 1- b 1- c
1
È 1 - 1 ˘
1
=
11.
(b - a) x ÍÎ1 - b 1 - ax ˙˚
(1 - ax)(1 - bx)
fi
a
=5
1+ r
a /(1 - r ) 20 1 + r
=4
=
fi
a /(1 + r ) 5
1- r
=
fi 1 + r = 4 – 4r
fi 5r = 3 fi r = 3/5.
7. Let f(x) = Ax2 + Bx + C
¥
2
\ f(x) = Ax + C
fi f¢(x) = 2Ax
Now, a, b, c are in A.P.
fi 2Aa, 2Ab, 2Ac are in A.P.
fi f ¢(a), f ¢(b), f ¢(c) are in A.P.
1
È(1 - bx)-1 - (1 - ax) -1 ˘˚
(b - a) x Î
1
1
xn in
is
(b - a) x
(1 - ax)(1 - bx)
n+1
n
x
in (1 – bx)
xn + 1 in (1 – ax)n]
As f(1) = f(–1), B = 0
1
1
and Tn =
m
n
1 2
n (n + 1)
2
fi 1 – a, 1 – b, 1 – c are in A.P.
2
a
a
6.
= 100
= 20 ,
1- r2
1- r
8. Tm =
+ (2m – 1)2
Putting n = 2m – 1, we get desired sum.
= Â x+ Â x- Â x
\
1
1 1
fid=
.
mn
n m
1
Thus, a - d = - md
n
1 1
= - =0
n n
9. Let n = 2m, then
( m - n) d =
= m(2m – 1)2
Required sum
x ŒS1
1
1
and a + (n - 1)d =
n
m
Subtracting we get
fi a + (m - 1)d =
=
1
{b n+1 - a n+1}
b-a
12. Let d be the common difference of the A.P., then
p
[2a + ( p - 1)d ] p 2
2 1
= 2
q
q
[2a1 + (q - 1)d ]
2
1
a1 + ( p - 1)d
p
2
fi
=
1
q
a1 + (q - 1)d
2
Progressions
Putting p = 11 and q = 41, we get
a6 11
=
a21 41
an are in H.P.
13. a1, a2,
fi
fi
fi
fi
1 1
1 are in A.P.
, ,
a1 a2
an
1 1
1 1
1
1
- = =
= d (say)
a2 a1 a3 a2
an an-1
1
1
a1a2 = (a1 - a2 ), a2 a3 = (a2 - a3 ) ,
d
d
1
an–1an = (a1 - an )
d
Thus,
a 1a 2 + a 2a 3 +
+ an –1an
1
[a1 - a2 + a2 - a3 +
+ a n – 1 – a n]
d
1
= (a1 - an )
d
1
1
a -a
But
= + (n - 1)d fi 1 n = (n - 1)d
an a1
an a1
=
\
a 1a 2 + a 2a 3 +
n
14. ar = ar
fi
n +1
+ ar
1=r+r
2
n+2
an
–1
an = (n – 1) a1an
, a > 0, r > 0
fi
2
r +r–1=0
1
r = ( 5 + 1) as r > 0
2
15. As n < n + 1 " n ≥ 2
fi
fi
n(n + 1) < (n + 1)2 " n ≥ 2
fi
n(n +1) < n + 1 " n ≥ 2
Thus, Statement-2 is true.
8.49
Subtracting, we get
2
1 4
4
4
S =1+ + 2 + 3 + 4 +
3
3 3
3 3
1 4 / 32
4 2
= 1+ +
= + =2
3 1 - 1/ 3
3 3
fi
S=3
17. Suppose he takes n minutes to count 4500 notes. We
have a1 + a2 +
+ a10 = 10(150), a11 = 148 and
a11, a12, ..., an is an A.P. with common difference d
= –2.
We are given.
a1 + a2 + ... + a10 + a11 + ... + an = 4500
fi
fi
fi
a11 + a12 +... + an = 3000
n - 10
[a11 + an ] = 3000
2
1
(n - 10)[148 + 148 + (n - 11)( -2)] = 3000
2
fi
(n – 10) (148 – n + 11) = 3000
fi
(n – 10) (159 – n ) = 3000
fi
n2 – 169n + 4590 = 0
fi
n2 – 135n – 34n + 4590 = 0
fi
(n – 135) (n – 34) = 0
fi
n = 135, 34
All the notes get counted in 34 minutes.
18. Suppose his total saving is ` 11040 after n months.
His savings are
200, 200, 200, 240, 280, ... upto n terms
Note that
Also, for n ≥ 2
1
1
1
+
+
+
n
1
2
200, 240, 280, ... upto (n – 2) terms is an A.P. whose
sum is 10640.
1
1
1
>
+
++
n
nn
Thus,
1
(n - 2)[2(200) + (n - 3)(40)] = 10640
2
fi (n – 2) (n + 7) = 532
1
1
1
+
+
+
> n
1
2
n
\ Statement-1 is true.
fi n2 + 5n – 546 = 0
fi n = – 26, 21
n times
fi
Thus, both the statements are correct but Statement-2
is not a correct explanation of Statement-1.
2 6 10 14
16. Let S = 1 + + 2 + 3 + 4 +
3 3 3 3
1
1 2 6 10
S= + 2+ 3+ 4+
3
3 3 3 3
As n > 0, n = 21.
19. Let d the common difference of the A.P. then
100
100
r =1
r=1
a – b = Â ( a2 r - a2 r -1 ) = Â d = 100d
fi d = 1 (a - b ) .
100
8.50
Complete Mathematics—JEE Main
20. We have
7
È179 + 10-20 ˘˚
81 Î
24. Sn = 2n + 3n2
=
100a100 = 500a50
fi
100 [a + 99d] = 50[a + 49d]
fi
(100 – 50)a = 50(49 – 2 ¥ 99)d
fi
a = – 149d
or
a150 = 0
fi
a = S1 = 2 + 3 = 5
a + (a + d) = S2 = 2(2) + 3(22) = 16
a + 149d = 0
\ d = 16 – 10 = 6
Sum of required A.P.
21. Statement-2 is true since
n
3
3
3
3
 (k 2 - (k - 1)3 ) = (1 – 0 ) + (2 – 1 ) +
k =1
fi
+
(n3 – (n – 1)3) = n3
n
 (k - (k - 1)) (k 2 + k (k - 1) + (k - 1)2 ) = n3
k =1
n
fi
 (k 2 + k (k - 1) + (k - 1)2 ) = n3
n
[2a + (n - 1)(2d )]
2
n
= [ 2(5) + (n - 1)(2)(6) ]
2
= 6n2 – n
=
25. ar =
k =1
Putting n = 20, we get
2r + 1
1 + 22 + ... + r 2
=
20
 ((k - 1)2 + k (k - 1) + k 2 ) = 203
k =1
fi
1 + (1 + 2 + 4) + (4 + 6 + 9) +
+
(361 + 380 + 400) = 8000
22. 2y = x + z
=
2
(
1
(2r + 1)
6 ) r ( r + 1)( 2r + 1)
6
1 ˆ
Ê1
= 6Á Ë r r + 1˜¯
r (r + 1)
11
1
11
\ Â ar = 6 Ê1 - ˆ =
Ë
¯
12
2
r=1
and 2 tan– 1 y = tan– 1x + tan– 1 z
fi
Ê 2y ˆ
–1 Ê x + z ˆ
tan Á
˜
ÁË
2 ˜ = tan
Ë1- y ¯
1 - xz ¯
fi
2y
x+ z
=
2
1 - xz
1- y
fi
y2 = xz
fi
1
2
( x + z )2 = xz fi (x – z) = 0
4
x=z
fi
–1
[
2y = x + z]
\ 2y = 2x fi y = x
fix=y=z
23. S = 0.7 + 0.77 + 0.777 +
+ upto 20 terms
=
7
[0.9 + 0.99 + 0.999 +
9
=
7
[20 – (0.1 + 0.01 + 0.001 +
9
(
7È
= Í20 9Í
Î
1
10
) (1 - ( 110)20 ) ˘
1 - 110
˙
˙
˚
upto 20 terms]
upto 20 terms)]
a/r, a, ar
and last three numbers be
a, ar, 2ar – a
We are given
ar – a = 6
and 2ar – a = a/r
fi 2r 2 – r = 1
fi (2r + 1) (r – 1) = 0
fi r = – 1/2, 1
r π 1 is not possible, for then a (1) – a = 6 or 0 =
16
Let r = –1/2.
Thus, a(–1/2) – a = 6 fi a = – 4
a/r = 8
\
27. 12 + 32 + 52 +
13
2
+ 252
13
= Â (2r - 1) = Â (4r 2 - 4r + 1)
r =1
r =1
4
Ê 1ˆ
= 4 Á ˜ (13)(14)(27) - (13)(14) + (13)
Ë 6¯
2
= 2925
Progressions
28. See Solution to Question 12.
10
10
= 10 Ê 11ˆ
Ë 10 ¯
k =1
k =1
k = 100
29. Â k (2k )2 = 4 Â k 3
4
(10) 2 (10 + 1)2 = 12100
4
1
ˆ
30. tk = tan– 1 ÊÁ
Ë 1 + (n + k - 1)(n + k ) ˜¯
10
11 10
- 10 - 10 Ê ˆ
Ë 10 ¯
9
[2a + (9 - 1)d ] = 9(a + 4d)
2
200 < 9(a + 4d) < 220
34. S9 =
As a + 4d is a positive integer
Ê (n + k ) - (n + k - 1) ˆ
= tan– 1 Á
Ë 1 + (n + k )(n + k - 1) ˜¯
a + 4d = 23, 24
= tan– 1 (n + k) – tan–1 (n + k – 1)
Also, a + d = 12
\ 3d = 11, 12
20
\ S = Â tk = tan– 1(n + 20) – tan–1 (n)
k =1
As d is an integer, d = 4
Thus, t4 = a + 3d = (a + 4d) – d = 20
20
Ê
ˆ
= tan– 1 Á
Ë 1 + n(n + 20) ˜¯
20
fi tan (S) = 2
n + 20n + 1
31. Let d be the common difference of the A.P., then
m = a4 – a7 + a10 = a4 + (a10 – a7)
[a + 4d = 23 is not possible]
35. From Question 25,
1 ˆ
Ê1
ar = 6 Ë r r + 1¯
20
120 k
1ˆ
Ê
=
fi  ar = 6 1 =
Ë 21¯
21 21
r=1
= a1 + 3d + 3d = a1 + 6d
\ k = 120
Now,
S13 = 13 [2a1 + (13 - 1)d ] = 13m
2
a a
, , a, ar, ar2
r2 r
Reciprocals are
32. Let three positive numbers in G.P. be
r2 r 1 1 1
, , , ,
a a a ar ar 2
a , a , ar, a > 0, r > 1
r
Now, 2(2a) = a + ar
r
fi r 2 – 4r + 1 = 0
We are given
fi a2 = 49 fi a = ± 7
fi r = 2 + 3 as r > 1.
2
Ê 11ˆ
Ê 11ˆ
33. k = 1 + 2 Á ˜ + 3 Á ˜ +
Ë 10 ¯
Ë 10 ¯
11
11
11 2
k=
+ 2 ÊÁ ˆ˜ +
Ë 10 ¯
10
10
Subtracting, we get
a / r 2 + a / r + a + ar + ar 2
= 49
r / a + r / a + 1/ a + 1/ ar + 1/ ar 2
2
9
11
11
+ 9 ÊÁ ˆ˜ + 10 ÊÁ ˆ˜
Ë 10 ¯
Ë 10 ¯
11 Ê 11ˆ 2
1
- k = 1+
+Á ˜ +
10
10 Ë 10 ¯
a + a = 35
r2
As r2 > 0, a = 7
Also,
11
+ 10 ÊÁ ˆ˜
Ë 10 ¯
10
11 9
11 10
+ ÊÁ ˆ˜ - 10 ÊÁ ˆ˜
Ë 10 ¯
Ë 10 ¯
and
\
1
1
2
+1 = 5 fi r =
2
4
r
2
a/r = 28
37. Two A.P. are
3, 7, 11, 15, 19, 23, 27, 31,
=
10
(1110)
-1
11 - 1
10
11
- 10 Ê ˆ
Ë 10 ¯
10
and
1, 6, 11, 16, 21, 26, 31,
Common terms are
8.51
8.52
Complete Mathematics—JEE Main
11, 31, 51,
6(2n - 1) 21
2n - 1 7
fi
=
=
n
4
2n
2
\ number of terms in the A.P. is 8.
fi
Thus, sum to 20 common terms is
20
[2(11) + (20 - 1)(20)] = 4020
2
38. G =
41.
ab ,
1 Ê 1 1ˆ
a+b
+
=
Ë
¯
2 a b
2ab
1
5
4
M
fi MG =
=
4
G 5
a
b 5
a+b 5
+
=
fi
= fi
b
a 2
2 ab 4
3n 2
fi 9n < 200
<
100 3
fi n < 200/9
fi n £ 22
Next,
M=
2 3n 5
£
<
3 100 3
fi
23 £ n £ 55.
n = 23
= 0 + Â n + 2(56)
n = 23
=
39. Let Sn denote the sum to n terms of the given geometric series, that is,
2 2
2
2
Sn = 1 - - 2 - 3 - n -1
3 3
3
3
1 n -1 ˘
È
3 ) Î1 - ( 3 )
˚
1 - 13
Ê 1ˆ
Now, Á ˜
Ë 3¯
55
55
\a:b=1:4
1
= 1-1+ Ê ˆ
Ë 3¯
n =1
n =1
1
a
= 2,
b
2
1
a
fi = 4,
4
b
= 1-
22
56
\ Â f (n) = Â f (n) + Â f (n) + f (56)
fi
(2
fi 200 £ 9n < 500
n -1
n -1
<
1
=Ê ˆ
Ë 3¯
33
(23 + 55) + 112 = 1399
2
42. m = (1 + n)/2
Also, l, G1, G2, G3, n are in G.P.
Let its common ratio be r, then
n = lr4 fi r4 = n/l.
Now, G41 + 2G42 + G43
= l 4 [r4 + 2(r2)4 + (r3)4]
n -1
È n 2n 2 n3 ˘
= l4 Í + 2 + 3 ˙
l
l ˚
Îl
= ln (l 2 + 2nl + n2)
1
100
= ln (l + n)2 = 4lm2n
fi 3n –1 > 100 fi n – 1 ≥ 5
fin≥6
Thus, least value of n is 6.
40. Let the A.P. be a, a + d, ..., a + (2n – 1)d
It is given
a + (a +2d) +
+ (a + (2n – 2)d) = 24(1)
and (a + d) + (a + 3d) + ...+ (a + (2n – 1)d) = 30 (2)
From (1) and (2), we get
d
+d+
+ d = 6 fi nd = 6
n times
Also,
21
1 21
a + (2n – 1)d – a = 10 =
=
2
2 2
21
(2n – 1)d =
2
43. tr =
13 + 23 + + r 3
1 + 3 + + (2r - 1)
r 2 (r + 1)2 / 4
r2
1
2
= (r + 1)
4
9
1 9
fi  tr =  (r + 1)2
4 r =1
r=1
=
=
1 È 10 2 ˘
 r - 1˙
4 ÍÎr =1
˚
=
1 È1
˘
(10)(10 + 1)(20 + 1) - 1˙
Í
4 Î6
˚
= 96
Progressions
30
44. Â (r + 2)(r - 3)
fi
r =16
15
3d
4d
=
[d = common difference of the A.P.]
a2
a5
fir=
= Â (r + 15 + 2)(r + 15 - 3)
r =1
4
a5
=
3
a2
48. Note that
3 2 1
4
1 , 2 , 3 , 4, 4 ,...
5 5 5
5
15
= Â (r + 17)(r + 12)
r =1
15
= Â (r 2 + 29r + 204)
a = 1
r =1
45. ar2 + ar3 = 60
4
5
Sum to 10 terms
10
= Â [a + (k - 1)d ]2
(1)
and (a)(ar)(ar2) = 1000
k =1
= 10a2 + 2ad(1 + 2 + … + 9) + d2(12 + 22 + … + 92)
Ê 1ˆ
= 10a2 + (ad)(9)(10) + d2 Á ˜ (9)(10)(19)
Ë 6¯
(2)
(2) fi (ar)3 = 103 fi ar = 10
\ (1) gives us
Ê 8ˆ
= 10 Á ˜
Ë 5¯
2
10r + 10r = 60 fi r + r – 6 = 0
fi (r + 3)(r – 2) = 0 fi r = – 3, 2.
49. For A.P.
ar + as = 2a1 + (r + s – 2)d
5
1
1
1 ˘
È1
+
= ÂÍ 2(n + 1) 2(n + 2) 6(n + 3) ˙˚
n =1 Î 6n
\ a3 + a15 + 2a1 + 16d = 2a9
Similarly a7 + a11 = 2a9.
5
1 1
1 ˆ 1Ê 1
1 ˆ
= Â ÈÍ Ê Ë
¯
Ë
n + 1 3 n + 1 n + 2¯
n =1 Î 6 n
1 1
1 ˆ˘
+ Ê
6 Ë n + 2 n + 3 ¯ ˙˚
1
1 1 1
1 1 1
k 1
fi = Ê1 - ˆ - Ê - ˆ + Ê - ˆ
Ë
¯
Ë
¯
3 6
6
3 2 7
6 Ë 3 8¯
=
5 È 1 1 1 ˘ 5 Ê 11 ˆ 55
- +
=
=
2 ÍÎ 6 7 24 ˙˚ 2 Ë 168 ¯ 336
47. Let an denotes the nth term of the A.P., then
a5 a9
=
= r (say)
a2 a5
fi
a5 - a2
a9 - a5
=
a2
a5
2
Ê 4ˆ
Ê 8ˆ Ê 4ˆ
+ Á ˜ Á ˜ (9)(10) + Á ˜ (15)(19)
Ë 5¯
Ë 5¯ Ë 5 ¯
16
16
[8 + 36 + 57] =
(101)
5
5
\ m = 101
Thus, t7 = ar6 = (ar)r5 = 10(25) = 320
5
1
46. k = Â
3 n=1 n(n + 1)(n + 2)(n + 3)
2
=
As a > 0, ar = 10, therefore, r = 2.
1 5
5 1 5
fi k= Ê ˆ- + Ê ˆ
2 Ë 6 ¯ 14 2 Ë 24 ¯
3
and common
5
difference d =
1
29
= (15)(16)(31) + (15)(16) + (204)(15)
6
2
= 7780
2
8.53
Now, 72 = 4a9 fi a9 = 18
17
[2a1 + (17 – 1)d]
2
= 17a9 = (17)(18) = 306
=
Previous Years' B-Architecture Entrance
Examination Questions
1.
1
, a, b are in G.P.
16 2
fi a = b/16
Also, as a, b, 1/6 are in H.P.
2(a)(1/ 6)
2a
=
b=
a + 1/ 6
6a + 1
From (1) and (2)
16a2 =
2a
6a + 1
(1)
(2)
8.54
Complete Mathematics—JEE Main
fi 8a (6a + 1) = 1
[
a π 0]
20
1
(1 + 2 +
k =1 k
= Â
2
fi 48a + 8a – 1 = 0
fi (12a – 1) (4a + 1) = 0
1 20
+ k ) = 2 Â ( k + 1)
k =1
1 1
◊ (20)(2 + 21) = 115
2 2
6. Required numbers are
=
fi a = 1/12, a = –1/4
When a = 1/12, b = 1/9
When a = –1/4, b = 1
203, 210,
2ab
a+b
Let common ratio of G.P. a, y, z, b
Now, 399 = 203 + 7 (n – 1)
2. x =
be r, then r3 = b/a and
\ yz = ab
and y3 + z3 = a3r3 +
Thus, required sum is
29
[203 + 399] = 8729
2
7. a + b + c = xb
b3
r3
fi b + b + br = xb , where r is common
r
ratio of the G.P and r π 1
3 Ê bˆ
3 Ê aˆ
= a Ë ¯ +b Ë ¯
a
b
= ab (a + b)
2ab
(ab)(a + b) = 2a2b2
a+b
ab
1
yz
\
=
=
2
3
3
2ab
2(ab)
x( y + z )
3. x = 1 + a + a2 +
1
= (x – 1)
r
1
1
But r + < -2 or r + > 2
r
r
fi r+
fi x(y3 + z3) =
=
1
1- a
1
1- b
fi a = 1 – 1/x, b = 1 – 1/y
y=
Now, 1 + ab + a2b2 + ...
1
1
=
=
1 - (1 - 1/ x)(1 - 1/ y )
1- ab
xy
=
x + y -1
4. 3b = a + b + c ≥ 3(abc)1/3
fi b ≥ 41/3= 22/3
20 1
1
1
1
5. Â i Ê +
+
+ ... + ˆ
Ë
i +1 i + 2
20 ¯
i=1 i
+
fi 57 = 29 + (n – 1) fi n = 29
=
y = ar, z = b/r
Ê1 1
= ÁË + +
1 2
, 399
1ˆ
˜
20 ¯
1ˆ
1ˆ
1
Ê1
Ê1 1
+2 Á + + + ˜ + 3 Á + + ˜ +20 Ê ˆ
Ë
¯
¯
Ë2 3
Ë 20 ¯
20
3
20
1
1
= (1)(1) + (1 + 2) + (1 + 2 + 3) +
2
3
1
+ (1 + 2 + + 20)
20
\ x – 1 < –2 or x – 1 > 2
fi x < – 1 or x > 3
fi x Œ ( – •, – 1) » (3, •)
8. Let two A.P.’s be
a, a + d, a + 2d, ...
and A, A + D, A + 2D, ...
We are given
n
[2a + (n - 1)d ] 3n + 8
2
=
n
[2 A + (n - 1) D ] 7n + 15
2
n -1
a+
d
3n + 8
2
fi
=
n -1
7
n + 15
A+
D
2
Put (n – 1)/2 = 11 or n = 23.
\
77
7
3(23) + 8
a + 11d
=
=
=
A + 11D 7(23) + 15 176 16
9. Let an denote the growth of height in the nth year.
We have
a1 = 50 cm, ar = 50 – 5(r – 1) = 55 – 5r
The tree ceases to grow in the rth year where ar = 0
fi r = 11
Progressions
So height of tree is
a1 + a2 +
+ a9
10
[50 + 5]cm = 275 cm = 2.75 m
2
10. rth term of the series is
=
tr =
=
49 - r
(r + 1)(r + 2)
50
51
r +1 r + 2
1
1 ˆ
1
= 50 Ê
Ë r + 1 r + 2¯ r + 2
1 1 1
, , are in A.P.
a b c
1 2 1 2r 1 2r - 1
fi = - =
- =
a b c c c
c
2 1 1
Also, - =
b a c
Also,
1
2a - b
=
c
ab
ab 2
c2
ab 2
\
=
=
2
(ab / c)
a
( 2a - b) 2
fi
= c(2r – 1) = e
n
48
48
1 ˆ
Ê 1 - 1 ˆ
-Â Ê
fi  tr = 50 Â
Ë
¯
Ë
r
+
r
+
1
2
r
+
2¯
r =1
r =1
r=1
48
48 1
1 1
= 50 Ê - ˆ - Â
Ë 2 50 ¯ r =1 r + 2
=
\K=–1
11. 2, 2x – 1, 2x + 3 are in G.P.
fi (2x – 1)2 = 2(2x + 3)
fi 22x – 2(2x) + 1 = 2(2x) + 6
fi 22x – 4(2x) – 5 = 0
fi (2x – 5) (2x + 1) = 0
[
k =1
2x + 1 > 0]
Thus, there is exactly one value of x.
12. Let r be the common ratio of the G.P. b, c, d, then
c
, c = c, d = cr
r
As c, d, e, are in A.P.
b=
e = 2d – c = 2cr – c = c(2r – 1)
2n
n +1
fi f(n) = Sn – Sn–1 =
=
51 50 Ê 1 ˆ
-Â
2 r =1Ë r ¯
fi 2x = 5
13. Let Sn = Â f(k ) =
fi
2n 2(n - 1)
n +1
n
2
2[n 2 - (n 2 - 1)]
=
n(n + 1)
(n + 1)n
1
n(n + 1)
=
f(n)
2
1 10
1
= Â (k 2 + k )
2 k =1
k =1 f( n)
10
fi Â
=
1 1
1 1
◊ (10)(11)(21) + ◊ (10)(11)
2 6
2 2
= 220
14. 2b = a + c,
2
1 1
= + , c2 = bd
d
c e
2ce
c2
=d=
b
c+e
e
c
c
fi
=
=
c+e
2b
a+c
Now,
fi ae + ce = c2 + ce
fi c2 = ae fi a, c, e are in G.P.
8.55
Limits and Continuity 9.1
CHAPTER NINE
LIMITS
Let f be a function defined in some neighbourhood of c, but
not necessarily at the point c. We are interested in the behaviour of f (x) when the distance between x and c is small, i.e.,
| x – c | is small. Specifically, we would like to know whether
there is some real number l such that f (x) approaches l as x
approaches c. We write
for each e > 0, there exists a d > 0 such that if c – d < x < c,
then | f (x) – l| < e.
Similarly we define, the right-hand limit written as
lim f (x) = l
xÆc+
Y
lim f (x) = l
l1
xÆc
to indicate that the difference between f (x) and l can be
made arbitrarily small (smaller than any pre-assigned positive number) simply by requiring that x be sufficiently close
to c but different from c.
Take a positive number e. limxÆ c f (x) = l, if | f (x) – l| < e
for all x sufficiently close to c but different from it. That is,
there exists a positive number d such that if 0 < | x – c | < d,
then | f (x) – l| < e.
y
l+e
l
l-e
a-d a a+d
x
Fig. 9.1
ONE-SIDED LIMITS
l2
X
c
lim f (x) = l1
x ® c–
lim f (x) = l2
x ® c+
Fig. 9.2
If limx Æ c + f (x) = lim x Æ c – f (x), then the limit of f (x) as
x Æ c exists and is written as limx Æ c f (x). The limit of a function f (x) may not exists in any of the following situations:
1. limx Æ c + f (x) does not exist, i.e., f (x) becomes infinitely small or large as x Æ c+. For example, as
x Æ 0 +, the function f (x) = 1/x becomes infinitely
large, as in this case x crosses over values which
are positive but sufficiently small.
2. limx Æ c – f (x) does not exist as for example, in
f(x) = 1/x as x Æ 0–, where x crosses over values
which are negative, so that f (x) becomes infinitely
small.
3. Both limx Æ c + f (x) and limx Æ c – f (x) exist but are
Ï 1, x > c
unequal. for example, in f ( x ) = Ì
In this
Ó-1, x < c
case, lim x Æ c+ f (x) = 1 and lim xÆc– f (x) = – 1.
The left-hand limit is written as
lim f (x) = l
xÆc-
If x approaches c from its left, then it crosses over values of
the form c – h, h > 0, the difference between f (x) and l can
be made arbitrarily small. More precisely, lim f ( x ) = l if
x Æ c-
x
f(x)
c + .001
1
c + .01
1
c + .1
1
c – .001
c – .01
c – .1
–1
–1
–1
9.2
Complete Mathematics—JEE Main
FREQUENTLY USED LIMITS
1. lim
xÆ0
sin x
tan x
= 1 = lim cos x = lim
xÆ0
xÆ0
x
x
sin -1 x
tan -1 x
= lim
.
xÆ0
xÆ0
x
x
1ˆ x
Ê
2. lim (1 + x)1/x = e = lim Á 1 + ˜ .
xÆ0
xÆ• Ë
x¯
1. If f, g are Polynomial, Trigonometric, Logarithmic or Exponential function such that g(a) π 0.
f ( x ) f (a )
=
provided f (a) and g(a) are
Then lim
x Æ a g (x)
g (a )
finite.
= lim
3. lim
xÆ0
log a (1 + x )
= loga e
x
(a > 0, a π 1).
xÆ0
log (1 + x )
= 1.
x
ax - 1
= log a (a > 0).
xÆ0
x
ex - 1
= log e = 1.
In particular, lim
xÆ0
x
(1 + x )m - 1
5. lim
= m, m ΠR.
xÆ0
x
log x
6. lim
= 0 (m > 0).
x Æ • xm
4. lim
xn - an
= nan – 1.
7. lim
xÆa x - a
SOME THEOREMS ON LIMITS
The calculation of limits is based on the following theorems.
If lim f (x) and lim g (x) exist, then
xÆc
1. lim (f (x) ± g (x)) = lim f (x) ± lim g (x).
xÆc
xÆc
xÆc
xÆc
x3 + 1
Since the numerator and denominator are polynomials and if f (x) = x2 + x + 2 and g(x) = x3 + 1 then
f(2) = 4 + 2 + 2 = 8 and g(2) = 8 + 1 = 9, thus
lim
xÆ2
f ( x ) f (2 ) 8
=
= .
g ( x ) g (2 ) 9
Illustration
3 + sin x
x Æ 0 cos x
Since the numerator and denominator are trigonometric function and if f(x) = 3 + sin x and g(x) =
cos x then f(0) = 3 and g(0) = 1, thus
3 + sin x 3 + 0
=
=3
x Æ 0 cos x
1
lim
2. If g(a) = 0 and f(a) = 0 and f, g are polynomials then
(x – a) is a factor of both f and g. In such a case we
factorise the numerator and denominator and cancel
out the common factor and then the situation is as
above.
Illustration
xÆc
xÆc
lim f ( x )
4. lim
xÆc
f ( x)
xÆc
=
(provided lim g(x) π 0).
g( x)
xÆc
lim g ( x )
xÆc
5. If f(x) > 0 and lim f ( x ) = L then L ≥ 0
xÆ a
For many elementary functions (e.g., polynomial, trigonometric, logarithmic and exponential), we have
lim f (x) = f Ê lim xˆ = f (c),
Ë xÆc ¯
in general, for continuous functions.
xÆc
COMPUTATION OF LIMITS
f (x)
Suppose we have find lim
x Æ a g (x)
2
Evaluate lim
3. lim (f (x) ◊ g (x)) = lim f (x) ◊ lim g(x).
xÆc
x2 + x + 2
xÆ2
2. lim [kf (x)] = k lim f(x) where k is a scalar.
xÆc
1
Evaluate lim
In particular taking a = e, lim
xÆc
Illustration
lim
x Æ1
3
x3 - 1
x2 - 1
If f(x) = x3 –1 and g(x) = x2 –1 then f(1) = g(1)
= 0. Therefore x –1 is a factor of x3 –1 and x2 –1,
hence
( x - 1) ( x 2 + x + 1)
x Æ1 x2 - 1
x Æ1
( x - 1) ( x + 1)
lim
x3 - 1
= lim
x2 + x + 1
x Æ1
x +1
1+1+1 3
=
= .
1+1
2
= lim
3. If f and g are of the form p ( x ) – q ( x ) and
r ( x ) - s( x ) where p(x), q(x), r(x), s(x) are polynomials such that f (a) = g(a) = 0 then we use rationalisation of the denominator and numerator, e.g.
Limits and Continuity 9.3
L’HÔPITAL’S RULES FOR
CALCULATING LIMITS
4
Illustration
4a + 3 x - x + 6a
lim
1. ( 0 0 form) If f and g are differentiable functions
on (0, d ) (see the definition of differentiability in
the next chapter) such that
(i) g¢(x) π 0 for any x Œ (0, d )
2 a + 5 x - 3a + 4 x
xÆa
( 2 x - 2 a ) ( 2 a + 5 x + 3a + 4 x )
= lim
( x - a)
xÆa
(
4a + 3 x + x + 6a
)
(ii)
(Rationalizing the numerator and denominator)
2a + 5 x + 3a + 4 x
= 2 lim
4a + 3 x + x + 6a
xÆa
4. lim
a
=
= 2.
-1
, where f(x) Æ 0 as x Æ 0
g (x)
f (x)
exist
g (x)
and lim
xÆ0
then lim
xÆ0
a f (x) - 1
a f (x) - 1 f ( x )
= lim
◊
xÆ0
g (x)
f (x) g (x)
xÆ0
f (x)
g (x)
5
Illustration
asin x - 1
x cos x
= lim
xÆ0
a
- 1 sin x 1
= log a.1.1 = log a
◊
x cos x
sin x
xÆ0
lim (1 + f(x))1/g(x) = lim (1 + f(x))1/f(x).( f (x) /g (x))
xÆ0
f ( x)
Ê f ¢( x ) ˆ
= L, then lim
= L.
(ii) lim Á
˜
x Æ • Ë g ¢( x ) ¯
x Æ • g( x )
3. (•/• form). If f and g are differentiable functions
on (0, d ) such that
(i) g¢(x) π 0 for any x Œ (0, d )
(ii) f (x) Æ •, g (x) Æ • as x Æ 0+
f ¢( x )
f ( x)
(iii) lim
= L, then lim
=L
g( x )
x Æ 0 + g ¢( x )
xÆ0+
7
What is wrong with the following application of L’Hôpital’s
rule?
lim
x3 + 3x - 4
2 x2 + x - 3
3x2 + 3
6x 3
= lim
= .
x Æ1 4 x + 1
x Æ1 4
2
= lim
3x2 + 3
0
is not of form as x Æ 1.
0
4x + 1
Therefore, it is not correct to apply L’Hôpital rule.
In fact
)
lim 3 x 2 + 3
3x2 + 3 x Æ1
6
lim
= .
=
x Æ1 4 x + 1
lim ( 4 x + 1) 5
6
x Æ1
To find lim (1 + (x2 + 3x))1/sinx, write the required limit as
xÆ0
(1 + ( x 2 + 3 x )[1 /( x
2
+ 3 x )] ( x ( x + 3) / sin x )
= lim (1 + (x2 + 3x))[(1/(x
2
4. The following illustrations will show the methods of
converting 0°, 1•, 0•, ••, • – • etc., to the forms
0
•
and
.
0
•
+ 3x)] (x(x +3) / sin x)
xÆ0
Illustration
2
= lim (1 + (x + 3x))
= e1.3 = e3
xƕ
(
= el
xÆ0
f ( x)
= L.
g( x )
The expression
xÆ0
lim
f ¢( x )
= L, then
g ¢( x )
2. (L¢ Hôpital’s rule when x Æ •). If f and g are differentiable functions on [a, •) such that
(i) lim f (x) = lim g (x) = 0
x Æ1
f (x)
lim
= l exist
x Æ 0 g (x)
then
xÆ0
lim
xÆ0+
Illustration
sin x
5. lim (1 + f (x))1/g(x), where f(x) Æ 0 as x Æ 0 and
Illustration
lim
xÆ0+
xÆ0+
xƕ
= log a lim
xÆ0
2 7a
(iii)
f (x)
xÆ0
lim
2 (2 7 a )
lim f (x) = 0 = lim g (x)
xÆ0+
8
(1/(x2 + 3x)) (x/sin x) (x + 3)
lim (sin x )
xÆ0 +
x
9.4
Complete Mathematics—JEE Main
It is of the form 00. If L is the limit sought, then
log sin x Ê • ˆ
log L = lim x log sin x = lim
Á ˜
x Æ 0+
x Æ 0+
1/ x Ë • ¯
Illustration
lim
0
- cos x
x cos x
= lim
=
=0
1
x
Æ
0
(sin x ) / x 1
sin x ÊÁ 2 ˆ˜
Ëx ¯
= lim
x Æ 0+
(1 + x )1 / x - e
x
Let y = (1 + x)1/x
xÆ0
fi log y =
Thus L = 1
Illustration
10
˘
x2 x3
1
1È
+
- ˙
log (1 + x) = Í x xÎ
x
2
3
˚
È
˘
x x2
= Í1 - +
- ˙
Î 2 3
˚
9
lim (1 – 3x)4/x (1• form)
xÆ0
fiy=
If L is the required limit then
È
˘
x x2
- ˙
Í1 - +
2
3
˚
eÎ
= e◊e
2
˘
ˆ
1 Ê x x2
˙
+
+
˜¯
2 ÁË 2 3
˙˚
Thus L = e–12
y-e
x
È 1 x
= e Í- + + terms containing
Î 2 3
USE OF SERIES EXPANSION IN
FINDING LIMITS
Thus
The following series expansions are useful in evaluating
limits
p ( p - 1) 2
p ( p - 1) ( p - 2 )
1. (1 + x)p = 1 + px +
x +
3!
2!
x3 + º
(for |x| < 1 and p ΠQ, the set of rational numbers)
x2 x3
+
+
2! 3!
2
3. ax = 1 + x (loge a) +
4. log (1 + x) = x –
x
(loge a)2 + º
2!
x2 x3 x 4
+
+ (|x| < 1)
2
3
4
Ê
ˆ
x2 x3
5. log (1 – x) = – Á x +
+
+ ˜ (|x| < 1)
2
3
Ë
¯
6. sin x = x -
x3 x 5
+
-
3! 5!
7. cos x = 1 -
x2 x4
+
-
2! 4!
8. tan x = x +
x3 2 x 5
+
+
3
15
9. sin–1 x = x +
10. tan–1 x = x -
x x2
+
-
2 3
È Ê x x2
ˆ
= e Í1 + Á - +
- ˜ +
¯
Î Ë 2 3
4 log (1 - 3 x ) Ê 0 ˆ
4 ( -3)
log L = lim
= – 12
ÁË ˜¯ = lim
xÆ0
x Æ 0 1 - 3x
x
0
2. ex = 1 + x +
-
1 x3 1 3 x 5
+ ◊
+ (–1< x <1)
2 3 2 4 5
x3 x 5
+
- (–1< x <1)
3
5
x 2 and higher powers of x˙˘
˚
\ lim
(1 + x )1 / x - e
x
xÆ0
=-
e
.
2
SEQUENTIAL LIMITS
A sequence is a function whose domain is the set of natural
numbers. If f: N Æ R then we write f (1) = f1, f (2) = f2 ,
f (n) = fn,
A sequence which has finite limit is said to be
convergent. Thus lim an = a if for any Π> 0 there exists a
nƕ
number N(depending on Œ) > 0 such that |an – a| < Œ holds
true for all n > N.
Illustration
1
1
are
, 2
nƕ
nƕ n
n
n
2n - 1
convergent sequences. If xn =
then lim
nƕ
2n + 1
xn = 1.
lim
1
n
11
= 0,
lim
1
2
= 0 so
lim f(x) = l if and only if for every sequence xn (of values
xÆa
of x) such that lim xn = a we have lim f(xn) = l.
n Æ•
nƕ
The above criterion is very useful in showing lim f(x) (for a
x Æa
Limits and Continuity 9.5
given f) does not exist. It is enough to give a sequence xn with
For x Π(0, 1), [x] = 0 so
lim xn = a such that lim f(xn) does not exist. Alternatively,
nƕ
nƕ
give two sequences xn and x¢n such that lim xn = a = lim x¢n
nƕ
Hence lim
nƕ
xÆ 0+
[ x]
=0
x
[ x]
= 0.
x
but lim f(xn) π lim f(x¢n).
nƕ
nƕ
CONTINUITY
12
Illustration
1
does not exist let f (x) =
x -1
1
2
1
sin
and choose xn = 1 +
, xn¢ = 1 +
,
(4n + 1)p
( x - 1)
np
so lim xn = 1 = lim x¢n. Now f (xn) = sin np = 0
To show that lim sin
xÆ1
nƕ
nƕ
4n + 1
p = 1. Hence
nƕ
2
lim f(x¢n) = 1. Thus lim f(x) does not exist.
so lim f(xn) = 0. f(xn) = sin
nƕ
nƕ
SOME USEFUL RESULTS ON LIMITS
In ordinary language, to say that a certain process is continuous implies that it goes on without interruption and without
abrupt changes. In mathematics, the word continuous has
much the same meaning. We shall first introduce the idea
of continuity at a point c, and then that of continuity on an
interval.
The function f is said to continuous at x = c if
limxÆ c f (x) = f (c) or, equivalently, limh Æ 0 f (c + h) = f (c).
The function f is said to be continuous on an open interval
(a, b) if it is continuous at each point of (a, b).
The function f is said to be continuous on a closed interval
[a, b] if
(i) f is continuous at each point of (a, b),
(ii) lim f (x) = f (a), (i.e., right hand side continuity
xÆa+
1. If lim f(x) = l with l π 0 and lim g(x) = 0 then
xÆc
lim
x Æc
at